7.2: Factoring Polynomials
Introduction
In this lesson you will learn to factor polynomials by removing the greatest common factor. You will also learn to factor polynomials of the form \begin{align*}ax^2+bx+c\end{align*} where \begin{align*}a=1\end{align*}. You will learn a method known as the ‘box’ method to do this factoring. You will use the same method to factor polynomials of the form \begin{align*}ax^2+bx+c\end{align*} where \begin{align*}a \neq 1\end{align*}. The last polynomials that you will earn to factor are those known as the sum and difference of two squares.
Objectives
The lesson objectives for Factoring Polynomials are:
- Factoring a common factor
- Factoring \begin{align*}ax^2+bx+c\end{align*} where \begin{align*}a=1\end{align*}
- Factoring \begin{align*}ax^2+bx+c\end{align*} where \begin{align*}a \neq 1\end{align*}
- Factoring the sum and difference of two squares
Factoring a Common Factor
Introduction
In this concept you will learn how to factor polynomials by looking for a common factor. You have used common factors before in earlier courses. Common factors are numbers (numerical coefficients) or letters (literal coefficients) that are a factor in all parts of the polynomials. In earlier courses of mathematics you would have studied common factors of two numbers. Say you had the numbers 25 and 35. Two of the factors of 25 are \begin{align*}{\color{red}5} \times {\color{red}5}\end{align*}. Two of the factors of 35 are \begin{align*}{\color{red}5} \times 7\end{align*}. Therefore a common factor of 25 and 35 would be \begin{align*}{\color{red}5}\end{align*}.
In this concept of Lesson Factoring Polynomials, you will be using the notion of common factors to factor polynomials. Sometimes simple polynomials can be factored by looking for a common factor among the terms in the polynomial. Often, for polynomials, this is referred to as the Greatest Common Factor. The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly) each of the terms of the polynomial. This concept will be the first explored in Lesson Factoring Polynomials.
Watch This
Khan Academy Factoring and the Distributive Property
Guidance
Factor the following polynomial: \begin{align*}12x^4 + 6x^3 + 3x^2\end{align*}.
Step 1: Identify the GCF of the polynomial
If you look at just the factors of the numbers you can see the following:
\begin{align*}& \quad 12x^4 + 6x^3 + 3x^2\\ & \ \swarrow \qquad \quad \downarrow \qquad \ \searrow\\ & 1 \times 12 \quad 1 \times 6 \quad 1 \times {\color{red}3}\\ & 2 \times 6 \quad \ \ 2 \times {\color{red}3}\\ & {\color{red}3} \times 4\end{align*}
Looking at the factors for each of the numbers, you can see that 12, 6, and 3 can all be divided by 3.
Also notice that each of the terms has an \begin{align*}x^2\end{align*} in common
\begin{align*}12x^4 &= 12 \cdot x \cdot x \cdot {\color{red}x} \cdot {\color{red}x}\\ 6x^3 &= 6 \cdot x \cdot {\color{red}x} \cdot {\color{red}x}\\ 3x^2 &= 3 \cdot {\color{red}x} \cdot {\color{red}x}\end{align*}
So the GCF for this polynomial is \begin{align*}3x^2\end{align*}
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}& 12x^4+6x^3+3x^2=3x^2(4x^2+2x+1)\\ & \qquad \qquad \qquad \qquad \qquad \swarrow \qquad \ \downarrow \qquad \ \searrow\\ & \qquad \qquad \text{Remember the} \ \ \text{Remember the} \ \text{Remember}\\ & \qquad \qquad \text{factors of} \ 12 \quad \ \ \text{factors of} \ 6 \qquad \text{the factors of}\\ & \qquad \qquad \text{included} \ {\color{red}3} \times 4 \quad \text{included} \ 2 \times {\color{red}3} \ \ 3 \ \text{are} \ 1 \times {\color{red}3}\\ & \qquad \qquad \text{and} \ {\color{red}x^2} \cdot x^2=x^4 \ \text{and} \ {\color{red}x^2} \cdot x=x^3 \ \text{and} \ {\color{red}x^2} \cdot 1=x^2\end{align*}
Example A
Factor the following binomial: \begin{align*}5a + 15\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 5a+15\\ & \ \ \swarrow \qquad \searrow\\ & 1 \times {\color{red}5} \qquad 1 \times 15\\ & \qquad \qquad \ 3 \times {\color{red}5}\end{align*}
Looking at the factors for each of the numbers, you can see that 5 and 15 can both be divided by 5.
So the GCF for this binomial is 5
Step 2: Divide out the GCF out of each term of the binomial
\begin{align*}5a + 15 = 5(a + 3)\end{align*}
Example B
Factor the following polynomial: \begin{align*}4x^2+8x-2\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad \ 4x^2+8x-2\\ & \ \ \swarrow \quad \quad \ \downarrow \quad \quad \searrow\\ & 1 \times 4 \quad 1 \times 8 \quad 1 \times {\color{red}2}\\ & 2 \times {\color{red}2} \quad {\color{red}2} \times 4\end{align*}
Looking at the factors for each of the numbers, you can see that 4, 8 and 2 can all be divided by 2.
So the GCF for this polynomial is 2
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}4x^2+8x-2=2(2x^2+4x-1)\end{align*}
Example C
Factor the following polynomial: \begin{align*}3x^5-9x^3-6x^2\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad \ 3x^5-9x^3-6x^2\\ & \ \ \swarrow \quad \quad \ \downarrow \qquad \ \searrow\\ & 1 \times {\color{red}3} \quad 1 \times 9 \quad 1 \times 6\\ & \qquad \quad \ 3 \times {\color{red}3} \quad 2 \times {\color{red}3}\end{align*}
Looking at the factors for each of the numbers, you can see that 3, 9 and 6 can all be divided by 3.
Also notice that each of the terms has an \begin{align*}x^2\end{align*} in common
\begin{align*}3x^5 &= 3 \cdot x \cdot x \cdot x \cdot {\color{red}x} \cdot {\color{red}x}\\ -9x^3 &= -9 \cdot x \cdot {\color{red}x} \cdot {\color{red}x}\\ -6x^2 &= -6 \cdot {\color{red}x} \cdot {\color{red}x}\end{align*}
So the GCF for this polynomial is \begin{align*}3x^2\end{align*}
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}3x^5-9x^3-6x^2=3x^2(x^3-3x-2)\end{align*}
Vocabulary
- Common Factor
- Common factors are numbers (numerical coefficients) or letters (literal coefficients) that are a factor in all parts of the polynomials.
- Greatest Common Factor
- The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly) each of the terms of the polynomial.
Guided Practice
- Find the common factors of the following: \begin{align*}a^2(b+7)-6(b+7)\end{align*}
- Factor the following polynomial: \begin{align*}5k^6+15k^4+10k^3+25k^2\end{align*}
- Factor the following polynomial: \begin{align*}27x^3y+18x^2y^2+9xy^3\end{align*}
Answers
1. \begin{align*}a^2(b+7)-6(b+7)\end{align*}
Step 1: Identify the GCF
\begin{align*}a^2(b+7)-6(b+7)\end{align*}
This problem is a little different in that if you look at the expression you notice that \begin{align*}(b + 7)\end{align*} is common in both terms. Therefore \begin{align*}(b + 7)\end{align*} is the common factor.
So the GCF for this expression is \begin{align*}(b + 7)\end{align*}
Step 2: Divide out the GCF out of each term of the expression
\begin{align*}a^2 (b+7)-6(b+7)=(a^2-6)(b+7)\end{align*}
2. \begin{align*}5k^6+15k^4+10k^3+25k^2\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 5k^6+15k^4+10k^3+25k^2\\ & \swarrow \qquad \quad \downarrow \qquad \quad \downarrow \qquad \quad \searrow\\ & 1 \times {\color{red}5} \quad 1 \times 15 \quad 1 \times 10 \quad 1 \times 25\\ & \qquad \quad \ 3 \times {\color{red}5} \quad \ 2 \times {\color{red}5} \quad \ \ 5 \times {\color{red}5}\end{align*}
Looking at the factors for each of the numbers, you can see that 5, 15, 10, and 25 can all be divided by 5.
Also notice that each of the terms has an \begin{align*}k^2\end{align*} in common
\begin{align*}5k^6 &= 5 \cdot k \cdot k \cdot k \cdot k \cdot {\color{red}k} \cdot {\color{red}k}\\ 15k^4 &= 15 \cdot k \cdot k \cdot {\color{red}k} \cdot {\color{red}k}\\ 10k^3 &= 10 \cdot k \cdot {\color{red}k} \cdot {\color{red}k}\\ 25k^2 &= 25 \cdot {\color{red}k} \cdot {\color{red}k}\end{align*}
So the GCF for this polynomial is \begin{align*}5k^2\end{align*}
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}5k^6+15k^4+10k^3+25k^2=5k^2(k^4+3k^2+2k+5)\end{align*}
3. \begin{align*}27x^3y+18x^2y^2+9xy^3\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 27x^3y+18x^2y^2+9xy^3\\ & \quad \swarrow \qquad \quad \ \downarrow \qquad \quad \ \searrow\\ & 1 \times 27 \quad \ \ 1 \times 18 \qquad \ 1 \times {\color{red}9}\\ & 3 \times {\color{red}9} \ \qquad 2 \times {\color{red}9} \ \ \qquad 3 \times 3\\ & \qquad \qquad \ \ 3 \times 6\end{align*}
Looking at the factors for each of the numbers, you can see that 27, 18 and 9 can all be divided by 9.
Also notice that each of the terms has an \begin{align*}xy\end{align*} in common
\begin{align*}27x^3y &= 27 \cdot x \cdot x \cdot {\color{red}x} \cdot {\color{red}y}\\ 18x^2y^2 &= 18\cdot x \cdot {\color{red}x} \cdot y \cdot {\color{red}y}\\ 9xy^3 &= 9 \cdot {\color{red}x} \cdot y \cdot y \cdot {\color{red}y}\end{align*}
So the GCF for this polynomial is \begin{align*}9xy\end{align*}
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}27x^3y+18x^2y^2+9xy^3=9xy(3x^2+2xy+y^2)\end{align*}
Summary
Remember that when you factor expressions such as polynomials by taking out a common factor, the first step is to find the greatest common factor for all of the terms in the polynomial. Once you have found this common factor, you must divide each of the terms of the polynomial by this common factor.
The common factor can be a number, a variable, or a combination of both. This means that you need to look at both the numbers in the terms of the polynomial and the variables of the polynomial.
Problem Set
Find the common factors of the following:
- \begin{align*}2x(x-5)+7(x-5)\end{align*}
- \begin{align*}4x(x-3)+5(x-3)\end{align*}
- \begin{align*}3x^2(e+4)-5(e+4)\end{align*}
- \begin{align*}8x^2(c-3)-7(c-3)\end{align*}
- \begin{align*}ax(x-b)+c(x-b)\end{align*}
Factor the following polynomial:
- \begin{align*}7x^2 + 14\end{align*}
- \begin{align*}9c^2+3\end{align*}
- \begin{align*}8a^2+4a\end{align*}
- \begin{align*}16x^2+24y^2\end{align*}
- \begin{align*}2x^2-12x+8\end{align*}
Factor the following polynomial:
- \begin{align*}32w^2x+16xy+8x^2\end{align*}
- \begin{align*}12abc+6bcd+24acd\end{align*}
- \begin{align*}15x^2y-10x^2y^2+25x^2y\end{align*}
- \begin{align*}12a^2b-18ab^2-24a^2b^2\end{align*}
- \begin{align*}4s^3t^2-16s^2t^3+12st^2-24st^3\end{align*}
Answers
Find the common factors...
- \begin{align*}2x(x-5)+7(x-5)\end{align*}
Step 1: Identify the GCF
\begin{align*}2x(x-5)+7(x-5)\end{align*}
Notice that \begin{align*}(x - 5)\end{align*} is common in both terms, therefore \begin{align*}(x - 5)\end{align*} is the common factor.
So the GCF for this expression is \begin{align*}(x - 5)\end{align*}
Step 2: Divide out the GCF out of each term of the expression
\begin{align*}2x(x-5)+7(x-5)=(2x+7)(x-5)\end{align*}
- \begin{align*}3x^2(e+4)-5(e+4)\end{align*}
Step 1: Identify the GCF
\begin{align*}3x^2(e+4)-5(e+4)\end{align*}
Notice that \begin{align*}(e + 4)\end{align*} is common in both terms, therefore \begin{align*}(e + 4)\end{align*} is the common factor.
So the GCF for this expression is \begin{align*}(e + 4)\end{align*}
Step 2: Divide out the GCF out of each term of the expression
\begin{align*}3x^2(e+4)-5(e+4)=(3x^2-5)(e+4)\end{align*}
- \begin{align*}ax(x-b)+c(x-b)\end{align*}
Step 1: Identify the GCF
\begin{align*}ax(x-b)+c(x-b)\end{align*}
Notice that \begin{align*}(x - b)\end{align*} is common in both terms, therefore \begin{align*}(x - b)\end{align*} is the common factor.
So the GCF for this expression is \begin{align*}(x - b)\end{align*}
Step 2: Divide out the GCF out of each term of the expression
\begin{align*}ax(x-b)+c(x-b)=(ax+c)(x-b)\end{align*}
Factor the following...
- \begin{align*}7x^2+14\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 7x^2+14\\ & \ \swarrow \qquad \ \ \searrow\\ & 1 \times {\color{red}7} \quad 1 \times 14\\ & \qquad \quad \ 2 \times {\color{red}7}\end{align*}
Looking at the factors for each of the numbers, you can see that 7 and 14 can both be divided by 7.
So the GCF for this polynomial is 7
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}7x^2+14=7(x^2+2)\end{align*}
- \begin{align*}8a^2+4a\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 8a^2+4a\\ & \ \swarrow \qquad \ \ \searrow\\ & 1 \times 8 \qquad 1 \times {\color{red}4}\\ & 2 \times {\color{red}4}\end{align*}
Looking at the factors for each of the numbers, you can see that 8 and 4 can both be divided by 4.
Also notice that each of the terms has an \begin{align*}a\end{align*} in common
\begin{align*}8a^2 &= 8 \cdot a \cdot {\color{red}a}\\ 4a &= 4 \cdot {\color{red}a}\end{align*}
So the GCF for this polynomial is \begin{align*}4a\end{align*}
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}8a^2+4a=4a(2a+1)\end{align*}
- \begin{align*}2x^2-12x+8\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad \ 2x^2-12x+8\\ & \ \swarrow \qquad \ \ \downarrow \qquad \ \searrow\\ & 1 \times {\color{red}2} \quad 1 \times 12 \quad 1 \times {\color{red}8}\\ & \qquad \quad \ {\color{red}2} \times 6 \quad \ {\color{red}2} \times 4\\ & \qquad \quad \ 3 \times 4\end{align*}
Looking at the factors for each of the numbers, you can see that 2, 12 and 8 can all be divided by 2.
So the GCF for this polynomial is 2.
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}2x^2-12x+8=2(x^2-6x+4)\end{align*}
Factor the following...
- \begin{align*}32w^2x+16xy+8x^2\end{align*}
Step 1: Identify the GCF
\begin{align*}& \quad 32w^2x+16xy+8x^2\\ & \ \quad \swarrow \quad \qquad \downarrow \qquad \ \ \searrow\\ & 1 \times 32 \qquad 1 \times 16 \quad \ \ 1 \times {\color{red}8}\\ & 2 \times 16 \qquad 2 \times {\color{red}8} \qquad 2 \times 4\\ & 4 \times {\color{red}8} \qquad \ 4 \times 4\end{align*}
Looking at the factors for each of the numbers, you can see that 32, 16 and 8 can all be divided by 8.
Also notice that each of the terms has an \begin{align*}x\end{align*} in common
\begin{align*}32w^2x &= 32 \cdot w \cdot w \cdot {\color{red}x}\\ 16xy &= 16 \cdot {\color{red}x} \cdot y\\ 8x^2 &= 8 \cdot x \cdot {\color{red}x}\end{align*}
So the GCF for this polynomial is \begin{align*}8x\end{align*}.
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}32w^2x+16xy+8x^2=8x(4w^2+2y+x)\end{align*}
- \begin{align*}15x^2y-10x^2y^2+25x^2y\end{align*}
Step 1: Identify the GCF
\begin{align*}& \ 15x^2y-10x^2y^2+25x^2y\\ & \ \swarrow \qquad \quad \ \downarrow \qquad \quad \ \searrow\\ & 1 \times 15 \quad 1 \times 10 \quad \ \ 1 \times 25\\ & 3 \times {\color{red}5} \quad \ \ 2 \times {\color{red}5} \qquad 5 \times {\color{red}5}\end{align*}
Looking at the factors for each of the numbers, you can see that 15, 10 and 25 can all be divided by 5.
Also notice that each of the terms has an \begin{align*}x^2y\end{align*} in common
\begin{align*}15x^2y &= 15 \cdot {\color{red}x} \cdot {\color{red}x} \cdot {\color{red}y}\\ -10x^2y^2 &= -10 \cdot {\color{red}x} \cdot {\color{red}x} \cdot {\color{red}y} \cdot y\\ 25x^2y &= 25 \cdot {\color{red}x} \cdot {\color{red}x} \cdot {\color{red}y}\end{align*}
So the GCF for this polynomial is \begin{align*}5x^2y\end{align*}.
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}15x^2y-10x^2y^2+25x^2y &= 5x^2y(3-2y+5)\\ &= 5x^2y(8-2y)\end{align*}
- \begin{align*}4s^3t^2-16s^2t^3+12st^2-24st^3\end{align*}
Step 1: Identify the GCF
\begin{align*}& 4s^3t^2-16s^2t^3+12st^2-24st^3\\ & \swarrow \qquad \quad \downarrow \qquad \quad \downarrow \qquad \quad \searrow\\ & 1 \times {\color{red}4} \quad 1 \times 16 \quad 1 \times 12 \quad \ \ 1 \times 24\\ & \qquad \quad \ 2 \times 8 \quad \ 2 \times 6 \qquad \ 2 \times 12\\ & \qquad \quad \ 4 \times {\color{red}4} \quad \ 3 \times {\color{red}4} \qquad \ 3 \times 8\\ & \qquad \qquad \qquad \quad \quad \qquad \quad \ {\color{red}4} \times 8\end{align*}
Looking at the factors for each of the numbers, you can see that 4, 16, 12, and 24 can all be divided by 4.
Also notice that each of the terms has an \begin{align*}st\end{align*} in common
\begin{align*}4s^3t^2 &= 4 \cdot s \cdot s \cdot {\color{red}s} \cdot t \cdot {\color{red}t}\\ -16s^2t^3 &= -16 \cdot s \cdot {\color{red}s} \cdot t \cdot t \cdot {\color{red}t}\\ 12st^2 &= 12 \cdot {\color{red}s} \cdot t \cdot {\color{red}t}\\ -24st^3 &= 12 \cdot {\color{red}s} \cdot t \cdot t \cdot {\color{red}t}\end{align*}
So the GCF for this polynomial is \begin{align*}4st\end{align*}.
Step 2: Divide out the GCF out of each term of the polynomial
\begin{align*}4s^3t^2-16s^2t^3+12st^2-24st^3=4st(s^2t-4st^2+3t-6t^2)\end{align*}
Factoring ax² + bx + c where a = 1
Introduction
When there are no common factors, you have to have an alternate method to factor polynomials. In this concept you will begin to factor polynomials that are known as quadratic expressions. A quadratic expression is one in which one variable will have an exponent of two and all other variables will have an exponent of one. The general form of a quadratic expression is \begin{align*}ax^2+bx+c\end{align*} where ‘\begin{align*}a\end{align*}’ and ‘\begin{align*}b\end{align*}’ are the coefficients of \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*}, respectively, and ‘\begin{align*}c\end{align*}’ is a constant. The term \begin{align*}ax^2\end{align*} is the one that is necessary to have a quadratic expression. These expressions are often the result of multiplying polynomials. In this concept, you will be working with quadratic expressions where the coefficient for ‘\begin{align*}a\end{align*}’ is equal to 1.
Quadratic expressions of the form \begin{align*}ax^2+bx+c\end{align*} can be factored using algebra tiles. To do this, the tiles modeling the trinomial are laid out. Next, with the \begin{align*}x^2\end{align*} tile in the upper left position the remaining tiles are arranged around it to form a rectangle. By forming a rectangle, you can easily see the factors of the quadratic expression. Remember you have used algebra tiles before in previous lessons and chapters. The large square tile represents an \begin{align*}x^2\end{align*}, the rectangular tile represents the \begin{align*}x\end{align*} tile and the small squares represent the 1's. These are shown for you below.
Remember as well that the reverse of these colored tiles are white and this represents the negatives.
Watch This
Khan Academy Factoring trinomials with a leading 1 coefficient
Guidance
An electronic sign is used to advertise specials outside a local store. It is shaped like a rectangle. If the dimensions of the sign are known to be represented by the trinomial \begin{align*}x^2-5x+6\end{align*}, find the binomials that represent the length and width of the electronic sign.
The first step to solving this problem is to lay out the tiles you need to represent the trinomial (or the quadratic expression).
The second step would be to arrange these tiles to form a rectangle.
The width of the rectangle is \begin{align*}(x - 2)\end{align*}
The length of the rectangle is \begin{align*}(x - 3)\end{align*}
Therefore the factors of \begin{align*}x^2-5x+6\end{align*} are \begin{align*}(x-3)(x-2)\end{align*} or \begin{align*}(x-2)(x-3)\end{align*}
Example A
Factor \begin{align*}x^2+5x+6\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
5 : \begin{align*}x\end{align*} tiles
6 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width of the rectangle is \begin{align*}(x + 2)\end{align*}
The length of the rectangle is \begin{align*}(x + 3)\end{align*}
Therefore the factors of \begin{align*}x^2+5x+6\end{align*} are \begin{align*}(x+3)(x+2)\end{align*} or \begin{align*}(x+2)(x+3)\end{align*}
Example B
Factor \begin{align*}x^2-3x+2\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
3 : \begin{align*}-x\end{align*} tiles
2 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x - 1)\end{align*}
The length of the rectangle is \begin{align*}(x - 2)\end{align*}
Therefore the factors of \begin{align*}x^2-3x+2\end{align*} are \begin{align*}(x-2)(x-1)\end{align*} or \begin{align*}(x-1)(x-2)\end{align*}
Example C
Factor \begin{align*}x^2-7x+12\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
7 : \begin{align*}-x\end{align*} tiles
12 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x - 3)\end{align*}
The length of the rectangle is \begin{align*}(x - 4)\end{align*}
Therefore the factors of \begin{align*}x^2-7x+12\end{align*} are \begin{align*}(x-3)(x-4)\end{align*} or \begin{align*}(x-4)(x-3)\end{align*}
Vocabulary
- Quadratic Expression
- A quadratic expression is a trinomial in which one variable will have an exponent of two and all other variables will have an exponent of one. The general form of a quadratic expression is \begin{align*}ax^2+bx+c\end{align*} where ‘\begin{align*}a\end{align*}’ and ‘\begin{align*}b\end{align*}’ are the coefficients of \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*}, respectively, and ‘\begin{align*}c\end{align*}’ is a constant.
Guided Practice
- Factor \begin{align*}a^2+6a+8\end{align*}.
- Factor \begin{align*}x^2-2x-8\end{align*}.
- Find the binomial dimensions of a rectangle that is represented by the trinomial \begin{align*}h^2+10h+16\end{align*}
Answers
1. Factor \begin{align*}a^2+6a+8\end{align*}.
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
6 : \begin{align*}+ x\end{align*} tiles
8 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(a + 2)\end{align*}
The length of the rectangle is \begin{align*}(a + 4)\end{align*}
Therefore the factors of \begin{align*}a^2+6a+8\end{align*}. are \begin{align*}(a+2)(a+4)\end{align*} or \begin{align*}(a+4)(a+2)\end{align*}
2. Factor \begin{align*}x^2-2x-8\end{align*}.
Step 1: Place the tiles that represent the trinomial in front of you. In this polynomial you need to find the factors of 8. You can’t build a rectangle with just 2 \begin{align*}-x\end{align*} tiles and 8 -1 tiles.
The factors of 8 are \begin{align*}1 \times 8\end{align*}, and \begin{align*}2 \times 4\end{align*}. Notice that if you use 4 \begin{align*}-x \end{align*} tiles and 2 \begin{align*}+x\end{align*} tiles you will end up with 2 \begin{align*}-x\end{align*} tiles left over. Try to build a rectangle with these.
1 : \begin{align*}x^2\end{align*} tile
4 : \begin{align*}-x\end{align*} tiles
2 : \begin{align*}+x\end{align*} tiles
8 : \begin{align*}-1\end{align*} tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x + 2)\end{align*}
The length of the rectangle is \begin{align*}(x - 4)\end{align*}
Therefore the factors of \begin{align*}x^2-2x-8\end{align*} are \begin{align*}(x+2)(x-4)\end{align*} or \begin{align*}(x-4)(x+2)\end{align*}
3. Find the binomial dimensions of a rectangle that is represented by the trinomial \begin{align*}h^2+10h+16\end{align*}.
The first step to solving this problem is to lay out the tiles you need to represent the trinomial.
The second step would be to arrange these tiles to form a rectangle.
The width of the rectangle is \begin{align*}(h + 2)\end{align*}
The length of the rectangle is \begin{align*}(h + 8)\end{align*}
Therefore the factors of \begin{align*}h^2+10h+16\end{align*} are \begin{align*}(h+8)(h+2)\end{align*} or \begin{align*}(h+2)(h+8)\end{align*}
Summary
In this concept you again used algebra tiles but you used them to facto trinomials known as quadratic expressions. These special trinomials were in the form of \begin{align*}ax^2 + bx + c\end{align*} where \begin{align*}a = 1\end{align*}. Algebra tiles are useful for factoring quadratic expressions because when you lay out the tiles you need and then form your rectangle, you have the binomial factors of the quadratic. Sometimes this is not useful, like when the constant \begin{align*}(c)\end{align*} is very large. In these cases, algebra tiles become very cumbersome to use. In the next few concepts, you will begin to explore other methods for factoring quadratics.
Problem Set
Factor the following trinomials.
- \begin{align*}x^2+5x+4\end{align*}
- \begin{align*}x^2+12x+20\end{align*}
- \begin{align*}a^2+13a+12\end{align*}
- \begin{align*}z^2+7z+10\end{align*}
- \begin{align*}w^2+8w+15\end{align*}
Factor the following quadratic expressions.
- \begin{align*}x^2-7x+10\end{align*}
- \begin{align*}x^2-10x+24\end{align*}
- \begin{align*}m^2-4m+3\end{align*}
- \begin{align*}s^2-6s+7\end{align*}
- \begin{align*}y^2-8y+12\end{align*}
Factor the following quadratic expressions.
- \begin{align*}b^2-x-12\end{align*}
- \begin{align*}x^2+x-12\end{align*}
- \begin{align*}x^2-5x-14\end{align*}
- \begin{align*}x^2-7x-44\end{align*}
- \begin{align*}y^2+y-20\end{align*}
Answers
Factor...
- \begin{align*}x^2+5x+4\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
5 : \begin{align*}+x\end{align*} tiles
4 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x + 1)\end{align*}
The length of the rectangle is \begin{align*}(x + 4)\end{align*}
Therefore the factors of \begin{align*}x^2+5x+4\end{align*} are \begin{align*}(x+2)(x+4)\end{align*} or \begin{align*}(x+4)(x+2)\end{align*}
- \begin{align*}a^2+13a+12\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
13 : \begin{align*}+ x\end{align*} tiles
12 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(a + 1)\end{align*}
The length of the rectangle is \begin{align*}(a + 12)\end{align*}
Therefore the factors of \begin{align*}a^2+13a+12\end{align*} are \begin{align*}(a+1)(a+12)\end{align*} or \begin{align*}(a+12)(a+1)\end{align*}
- \begin{align*}w^2+8w+15\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
8 : \begin{align*}+ x\end{align*} tiles
15 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(w + 3)\end{align*}
The length of the rectangle is \begin{align*}(w + 5)\end{align*}
Therefore the factors of \begin{align*}w^2+8w+15\end{align*} are \begin{align*}(w+3)(w+5)\end{align*} or \begin{align*}(w+5)(w+3)\end{align*}
Factor...
- \begin{align*}x^2-7x+10\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
7 : \begin{align*}- x\end{align*} tiles
10 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x - 2)\end{align*}
The length of the rectangle is \begin{align*}(x - 5)\end{align*}
Therefore the factors of \begin{align*}x^2-7x+10\end{align*} are \begin{align*}(x-2)(x-5)\end{align*} or \begin{align*}(x-5)(x-2)\end{align*}
- \begin{align*}m^2-4m+3\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2 \end{align*} tile
4 : \begin{align*}- x \end{align*} tiles
3 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(m - 1)\end{align*}
The length of the rectangle is \begin{align*}(m - 3)\end{align*}
Therefore the factors of \begin{align*}m^2-4m+3\end{align*} are \begin{align*}(m-1)(m-3)\end{align*} or \begin{align*}(m-3)(m-1)\end{align*}
- \begin{align*}y^2-8y+12\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you.
1 : \begin{align*}x^2\end{align*} tile
8 : \begin{align*}- x\end{align*} tiles
12 : 1 tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(y - 2)\end{align*}
The length of the rectangle is \begin{align*}(y - 6)\end{align*}
Therefore the factors of \begin{align*}y^2-8y+12\end{align*} are \begin{align*}(y-2)(y-6)\end{align*} or \begin{align*}(y-6)(y-2)\end{align*}
Factor...
- \begin{align*}b^2-x-12\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you. Remember, in this polynomial you need to find the factors of 12. You can’t build a rectangle with just one \begin{align*}-x\end{align*} tiles and twelve -1 tiles.
Factors of 12
\begin{align*}& 1 \times 12\\ & 2 \times 6\\ & 3 \times 4\end{align*}
Notice that if you use 4 \begin{align*}-x\end{align*} tiles and 3 \begin{align*}+x\end{align*} tiles you will end up with one \begin{align*}-x\end{align*} tiles left over. Try to build a rectangle with these.
1 : \begin{align*}x^2\end{align*} tiles
4 : \begin{align*}- x\end{align*} tiles
3 : \begin{align*}+ x\end{align*} tiles
12 : \begin{align*}-1\end{align*} tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x + 3)\end{align*}
The length of the rectangle is \begin{align*}(x - 4)\end{align*}
Therefore the factors of \begin{align*}b^2-x-12\end{align*} are \begin{align*}(x+3)(x-4)\end{align*} or \begin{align*}(x-4)(x+3)\end{align*}
- \begin{align*}x^2-5x-14\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you. Remember, in this polynomial you need to find the factors of 14. You can’t build a rectangle with just five \begin{align*}-x\end{align*} tiles and fourteen -1 tiles.
Factors of 14
\begin{align*}& 1 \times 14\\ & 2 \times 7\end{align*}
Notice that if you use 5 \begin{align*}-x\end{align*} tiles and 2 \begin{align*}+x\end{align*} tiles you will end up with five \begin{align*}-x\end{align*} tiles left over. Try to build a rectangle with these.
1 : \begin{align*}x^2\end{align*} tile
7 : \begin{align*}- x\end{align*} tiles
2 : \begin{align*}+ x\end{align*} tiles
14 : \begin{align*}-1\end{align*} tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(x + 2)\end{align*}
The length of the rectangle is \begin{align*}(x - 7)\end{align*}
Therefore the factors of \begin{align*}x^2-5x-14\end{align*} are \begin{align*}(x+2)(x-7)\end{align*} or \begin{align*}(x-7)(x+2)\end{align*}
- \begin{align*}y^2+y-20\end{align*}
Step 1: Place the tiles that represent the trinomial in front of you. Remember, in this polynomial you need to find the factors of 20. You can’t build a rectangle with just one \begin{align*}+x\end{align*} tiles and twenty -1 tiles.
Factors of 20
\begin{align*}& 1 \times 20\\ & 2 \times 10\\ & 4 \times 5\end{align*}
Notice that if you use 5 \begin{align*}+x\end{align*} tiles and 4 \begin{align*}-x\end{align*} tiles you will end up with one \begin{align*}+x\end{align*} tiles left over. Try to build a rectangle with these.
1 : \begin{align*}x^2\end{align*} tile
5 : \begin{align*}+ x\end{align*} tiles
4 : \begin{align*}- x\end{align*} tiles
20 : \begin{align*}-1\end{align*} tiles
Step 2: Now, form the rectangular using the tiles.
The width if the rectangle is \begin{align*}(y + 5)\end{align*}
The length of the rectangle is \begin{align*}(y - 4)\end{align*}
Therefore the factors of \begin{align*}y^2+y-20\end{align*} are \begin{align*}(y+5)(y-4)\end{align*} or \begin{align*}(y-4)(y+5)\end{align*}
Factoring ax² + bx + c where a ≠ 1
Introduction
In the last concept you learned how to factor quadratics or trinomials \begin{align*}(ax^2+bx+c)\end{align*} where \begin{align*}a = 1\end{align*}. You learned that algebra tiles were very useful as a tool for solving these problems and that by building rectangles, finding the factors of quadratics was a fast way to find the solution. In this lesson, you will work with quadratics, or trinomials where the value of a does not equal 1. In these problems, algebra tiles may still be useful but may, at some point, become cumbersome due to the large numbers of tiles necessary to build the rectangles. You will use algebra to solve these problems. There are a few other methods that are useful for solving quadratics where \begin{align*}a \neq 1\end{align*}. One method you learned earlier is the FOIL pattern. Remember that with FOIL, you multiply the First two terms, the Outside terms, the Inside terms and the Last terms. So for example, with the trinomial:
\begin{align*}(2x-5)(x+7)=2x^2+9x-35\end{align*}
When you factor trinomials, you are really just using \begin{align*}{\color{red}\mathbf{F}}\mathbf{OI}{\color{blue}\mathbf{L}}\end{align*} in reverse. Let’s look at the example below. With the trinomial above \begin{align*}(2x^2+9x-35)\end{align*}, you had to factor both the 2 and the -35 (both the first number and the last number). You can say then, in general terms, that with the trinomial \begin{align*}ax^2+bx+c\end{align*}, you have to factor both “\begin{align*}a\end{align*}” and “\begin{align*}c\end{align*}”.
\begin{align*}& ax^2+bx+c=({\color{red}d}x+{\color{blue}e})({\color{red}f}x+{\color{blue}g})\\ & \text{where} \ a={\color{red}d} \times {\color{red}f} \ \text{and} \ c={\color{blue}e} \times {\color{blue}g}\end{align*}
The middle term \begin{align*}(b)\end{align*} is the sum of the outside two and the inside two terms. Therefore
\begin{align*}b = {\color{red}d} {\color{blue}g} + {\color{blue}e} {\color{red}f}\end{align*}
In this lesson you will have the opportunity to work through a number of examples to develop mastery at factoring trinomials of the form \begin{align*}ax^2+bx+c\end{align*} where \begin{align*}a \neq 1\end{align*} using this reverse FOIL method. Now let's begin.
Guidance
Jack wants to construct a border around his garden. The garden measures 5 yards by 18 yards. He has enough stone to build a border with a total area of 39 square yards. The boarder will be twice as wide on the shorter end. What are the dimensions of the border?
\begin{align*}\text{Area of Garden} &= 18 \times 5 = 90 \ yd^2\\ \text{Area of border} &= 39 \ yd^2\\ \text{Area of Garden} + \text{border} &= (12 + 2x)(5 + x)\\ \text{Area of border} &= (\text{Area of garden} + \text{border}) - \text{Area of garden}\\ 39 &= (18 + x)(5 + x) - 90\\ 39 &= 90+18x+5x+2x^2-90\\ 39 &= 23x+2x^2\\ 0 &= 2x^2+23x-39\end{align*}
Remember: \begin{align*}ax^2+bx+c=({\color{red}d}x+{\color{blue}e})({\color{red}f}x+{\color{blue}g})\end{align*}
To factor this trinomial try this method:
In this trinomial, the ‘a’ value is 2 and the ‘c’ value is -39. These values are placed in a box
As shown in the illustration below
The product of 2 and -39 is -78. Find the factors for -78 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of +23.
\begin{align*}& 1 \qquad -78 \qquad -1 \qquad 74\\ & 2 \qquad -39 \qquad -2 \qquad 39\\ & 3 \qquad -26 \qquad -3 \qquad 26\\ & 6 \qquad -13 \qquad -6 \qquad 13\end{align*}
From the list, -3 and 26 will work. Put the factors in the box as shown.
Remember the Greatest common factor from the first part of this Lesson. Going across the horizontal rows in the box, find the GCF of 2 and 26 it will be 2. Find the GCF of -3 and -39 it will be -3. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 2 and -3 it will be 1. Find the GCF of 26 and -39 it will be 13.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(1x + 13) (2x - 3)\end{align*}
So \begin{align*}2x^2+23x-39=(1x+13)(2x-3)\end{align*}. To find the dimensions of the border:
\begin{align*}& \ (x+13)(2x-3)=0\\ & \ \swarrow \qquad \qquad \searrow\\ & x+13=0 \ \ 2x-3=0\\ & x=-13 \quad \ \ x=\frac{3}{2}\end{align*}
Since \begin{align*}x\end{align*} cannot be negative, \begin{align*}x\end{align*} must equal \begin{align*}\frac{3}{2}\end{align*}.
Back to the question!
Width of Border: \begin{align*}2x = 2 \left(\frac{3}{2}\right) = 3 \ yd\end{align*}
Length of Border: \begin{align*}x = \frac{3}{2} \ yd\end{align*}
Example A
Factor: \begin{align*}2x^2+11x+15\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 2 and the ‘\begin{align*}c\end{align*}’ value is 15. Place these values in a box.
The product of 2 and 15 is 30. Next find the factors for 30 then look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of +11.
\begin{align*}& 1 \quad 30\\ & 2 \quad 15\\ & 3 \quad 10\\ & 5 \quad 6\end{align*}
From the list, 5 and 6 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 2 and 6. It will be 2. Find the GCF of 5 and 15. It will be 5. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 2 and 5. It will be 1. Find the GCF of 6 and 15. It will be 3.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(1x + 3) (2x + 5)\end{align*}
So \begin{align*}2x^2+11x+15=(x+3)(2x+5)\end{align*}
Example B
Factor: \begin{align*}3x^2-8x-3\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 3 and the ‘\begin{align*}c\end{align*}’ value is -3. Place the values in a box.
The product of 3 and -3 is -9. Find the factors for -9 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of -8.
\begin{align*}& 1 \qquad -9 \qquad -1 \qquad 9\\ & 3 \qquad -3\end{align*}
From the list, 1 and -9 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 3 and 1. It will be 1. Find the GCF of -9 and -3. It will be -3. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 3 and -9. It will be 3. Find the GCF of 1 and -3. It will be 1.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(3x + 1) (1x - 3)\end{align*}
So \begin{align*}3x^2-8x-3=(3x+1)(x-3)\end{align*}
Example C
Factor: \begin{align*}5w^2-21w+18\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 5 and the ‘\begin{align*}c\end{align*}’ value is 18. Place the values in a box.
The product of 5 and 18 is 90. Find the factors for 90 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of -21.
\begin{align*}& 1 \qquad 90 \qquad -1 \qquad -90\\ & 2 \qquad 45 \qquad -2 \qquad -45\\ & 3 \qquad 30 \qquad -3 \qquad -30\\ & 5 \qquad 18 \qquad -5 \qquad -18\\ & 6 \qquad 15 \qquad -6 \qquad -15\\ & 9 \qquad 10 \qquad -9 \qquad -10\end{align*}
From the list, -6 and -15 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 5 and -6. It will be 1. Find the GCF of -15 and 18. It will be 3. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 5 and -15. It will be 5. Find the GCF of -6 and 18. It will be 6.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(5x - 6) (x - 3)\end{align*}
So \begin{align*}5w^2-21w+18=(5x-6)(x-3)\end{align*}
Vocabulary
- Greatest Common Factor
- The Greatest Common Factor (or GCF) is the largest monomial that is a factor of (or divides into evenly) each of the terms of the polynomial.
- Quadratic Expression
- A quadratic expression is a trinomial in which one variable will have an exponent of two and all other variables will have an exponent of one. The general form of a quadratic expression is \begin{align*}ax^2 + bx + c\end{align*} where ‘\begin{align*}a\end{align*}’ and ‘\begin{align*}b\end{align*}’ are the coefficients of \begin{align*}x^2\end{align*} and \begin{align*}x\end{align*}, respectively, and ‘\begin{align*}c\end{align*}’ is a constant.
Guided Practice
- Factor the following trinomial: \begin{align*}8c^2-2c-3\end{align*}
- Factor the following trinomial: \begin{align*}3m^2+3m-60\end{align*}
- Factor the following trinomial: \begin{align*}5e^3+30e^2+40e\end{align*}
Answers
Factor...
1. \begin{align*}8c^2-2c-3\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 8 and the ‘\begin{align*}c\end{align*}’ value is -3. Place the values in a box.
The product of 8 and -3 is -24. Next, find the factors for -24 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of -2.
\begin{align*}& 1 \qquad -24 \qquad -1 \qquad 24\\ & 2 \qquad -12 \qquad -2 \qquad 12\\ & 3 \qquad \ -8 \qquad -3 \qquad \ \ 8\\ & 4 \qquad \ -6 \qquad -4 \qquad \ \ 6\end{align*}
From the list, 4 and -6 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 8 and 4. It will be 4. Find the GCF of -6 and -3. It will be -3. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 8 and -6. It will be 2. Find the GCF of 4 and -3. It will be 1.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(2c+1)(4c-3)\end{align*}
So \begin{align*}8c^2-2c-3=(2c+1)(4c-3)\end{align*}
2. \begin{align*}3m^2+3m-60\end{align*}
First you can factor out the 3 from the polynomial
\begin{align*}3m^2+3m-60=3(m^2+m-20)\end{align*}
In this simpler trinomial, the ‘\begin{align*}a\end{align*}’ value is 1 and the ‘\begin{align*}c\end{align*}’ value is -20. Place the values in a box.
The product of 1 and -20 is -20. Next, find the factors for -20 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of 1.
\begin{align*}& 1 \qquad -20 \qquad -1 \qquad 20\\ & 2 \qquad -10 \qquad -2 \qquad 10\\ & 4 \qquad \ -5 \qquad -4 \qquad \ \ 5\end{align*}
From the list, -4 and 5 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 1 and -4. It will be 1. Find the GCF of 5 and -20. It will be 5. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 1 and 5. It will be 1. Find the GCF of -4 and -20. It will be -4.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(1m-4) (1m+5)\end{align*}
So \begin{align*}3m^2+3m-20=3(m-4)(m+5)\end{align*}
3. \begin{align*}5e^3+30e^2+40e\end{align*}
First you can factor out the 5 and one \begin{align*}e\end{align*} from the polynomial
\begin{align*}5e^3+30e^2+40e=5e(e^2+6e+8)\end{align*}
First, use the box for the simpler trinomial to factor it:
The product of 1 and 8 is 8.
Need: factors for 8 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 6.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(1e+2)(1e+4)\end{align*}
So \begin{align*}5e^3+30e^2+40e=5e(e+2)(e+4)\end{align*}
Summary
In the first lesson of this section you learned how to take a common factor out of a polynomial. In the next section you learned to factor trinomials (or quadratics) of the form \begin{align*}ax^2 + bx + c\end{align*} where \begin{align*}a = 1\end{align*} using algebra tiles. In this lesson, you continued working with trinomials (or quadratics) but in these quadratics \begin{align*}a \neq 1\end{align*}. As well, in this lesson, you were introduced to a neat method known as the box method. In this method, any quadratic can be solved by finding the factors that multiply together to give “\begin{align*}ac\end{align*}” and combine to give “\begin{align*}b\end{align*}”. Using the box method and finding GCFs, the factors of the quadratic can be found.
Problem Set
Factor the following trinomials
- \begin{align*}13x^2+5x+2\end{align*}
- \begin{align*}5x^2+9x-2\end{align*}
- \begin{align*}4x^2+x-3\end{align*}
- \begin{align*}2x^2+7x+3\end{align*}
- \begin{align*}2y^2-15y-8\end{align*}
Factor the following trinomials
- \begin{align*}2x^2-5x-12\end{align*}
- \begin{align*}2x^2+11x+12\end{align*}
- \begin{align*}6w^2-7w-20\end{align*}
- \begin{align*}12w^2+13w-35\end{align*}
- \begin{align*}3w^2+16w+21\end{align*}
Factor the following trinomials
- \begin{align*}16a^2-18a-9\end{align*}
- \begin{align*}36a^2-7a-15\end{align*}
- \begin{align*}15a^2+26a+8\end{align*}
- \begin{align*}20m^2+11m-4\end{align*}
- \begin{align*}3p^2+17p-20\end{align*}
Answers
Factor...
- \begin{align*}3x^2+5x+2\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 3 and the ‘\begin{align*}c\end{align*}’ value is 2. Place the values in a box.
The product of 3 and 2 is 6. Next, find the factors for 6 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of 5.
\begin{align*}& 2 \quad 6\\ & 3 \quad 3\end{align*}
From the list, 2 and 3 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 3 and 2. It will be 1. Find the GCF of 3 and 2. It will be 1. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 3 and 3. It will be 3. Find the GCF of 2 and 2. It will be 2.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(3x+2)(1x+1)\end{align*}
So \begin{align*}3x^2+5x+2=(3x+2)(x+1)\end{align*}
- \begin{align*}4x^2+x-3\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 4 and the ‘\begin{align*}c\end{align*}’ value is -3. Place the values in a box.
The product of 4 and -3 is -12. Next, find the factors for -12 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of 1.
\begin{align*}& 3 \qquad -12 \qquad -1 \qquad 12\\ & 4 \qquad \ -6 \qquad -2 \qquad \ \ 6\\ & 4 \qquad \ -4 \qquad -3 \qquad \ \ 4\end{align*}
From the list, -3 and 4 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 4 and -3. It will be 1. Find the GCF of 4 and -3. It will be 1. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 4 and 4. It will be 4. Find the GCF of -3 and -3. It will be -3.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(4x-3)(1x+1)\end{align*}
GCF for vertical rows
So \begin{align*}4x^2+x-3=(4x-3)(x+1)\end{align*}
- \begin{align*}2y^2-15y-8\end{align*}
First, the box:
The product of 2 and -8 is -16.
Need: factors for -16 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -15.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(2y+1)(1y-8)\end{align*}
So \begin{align*}2y^2-15y-8=(2y+1)(y-8)\end{align*}
Factor...
- \begin{align*}2x^2-5x-12\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 2 and the ‘\begin{align*}c\end{align*}’ value is -12. Place the values in a box.
The product of 2 and -12 is -24. Find the factors for -24 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of -5.
\begin{align*}& 1 \qquad -24 \qquad -1 \qquad 24\\ & 2 \qquad-12 \qquad -2 \qquad 12\\ & 3 \qquad \ -8 \qquad -3 \qquad \ \ 8\\ & 4 \qquad \ -6 \qquad -4 \qquad \ \ 6\end{align*}
From the list, 3 and -8 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 2 and 3. It will be 1. Find the GCF of -8 and -12. It will be -4. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 2 and -8. It will be 2. Find the GCF of 3 and -12. It will be 3.Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(2x+3)(1x-4)\end{align*}
So \begin{align*}2x^2-5x-12=(2x+3)(x-4)\end{align*}
- \begin{align*}6w^2-7w-20\end{align*}
First, the box:
The product of 2 and -20 is -120.
Need: factors for -120 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -7.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(2w-5)(3w+4)\end{align*}
So \begin{align*}6w^2-7w-20=(2w-5)(3w+4)\end{align*}
- \begin{align*}3w^2+16w+21\end{align*}
First, the box:
The product of 3 and 21 is 63.
Need: factors for 63 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 16.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(3w+7)(1w+3)\end{align*}
So \begin{align*}3w^2+16w+21=(3w+7)(w+3)\end{align*}
Factor...
- \begin{align*}16a^2-18a-9\end{align*}
In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 16 and the ‘\begin{align*}c\end{align*}’ value is -9. Place the values in a box.
The product of 16 and -9 is -9. Find the factors for -144 and look for the factor pair that will combine (add or subtract) to give the ‘\begin{align*}b\end{align*}’ value of -18.
\begin{align*}& 1 \qquad -144 \qquad -1 \qquad 9\\ & 2 \qquad -72 \qquad -2 \qquad 72\\ & 3 \qquad -48 \qquad -3 \qquad 48\\ & 4 \qquad -36 \qquad -4 \qquad 36\\ & 6 \qquad -24 \qquad -6 \qquad 24\\ & 8 \qquad -18 \qquad -8 \qquad 18\\ & 9 \qquad -16 \qquad -9 \qquad 16\\ & 12 \quad \ \ -12\end{align*}
From the list, 6 and -24 will work. Put the factors in the box as shown.
Going across the horizontal rows in the box, find the GCF of 16 and 6. It will be 2. Find the GCF of -24 and -9. It will be -3. Put these numbers in the box as shown.
THEN - going down the vertical rows...
Find the GCF of 16 and -24. It will be 8. Find the GCF of 6 and -9. It will be 3.
Put these numbers in the box as shown.
These new numbers represent your factors. \begin{align*}(8a+3)(2a-3)\end{align*}
So \begin{align*}16a^2-18a-9=(8a+3)(2a-3)\end{align*}
- \begin{align*}15a^2+26a+8\end{align*}
First, the box:
The product of 15 and 8 is 120.
Need: factors for 120 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 26.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(5a+2)(3a+4)\end{align*}
So \begin{align*}15a^2+26a+8=(5a+2)(3a+4)\end{align*}
- \begin{align*}3p^2+17p-20\end{align*}
First, the box:
The product of 3 and -20 is -60.
Need: factors for -16 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 17.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(3p+20)(1p-1)\end{align*}
So \begin{align*}3p^2+17p-20=(3p+20)(p-1)\end{align*}
Factoring the Sum and Difference of Two Squares
Introduction
In lesson Addition and Subtraction of Polynomials you learned about three special cases where binomials can be expanded using the distributive property to make polynomials. These special cases were
- \begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}
- \begin{align*}(x-y)^2=x^2-2xy+y^2\end{align*}
- \begin{align*}(x+y)(x-y)=x^2-y^2\end{align*}
In this lesson, you will again be working with these special cases. The first two special cases are the sum of two squares. These are often called perfect square trinomials. The third special case is called the difference of two squares. Rather than expanding the binomials to make polynomials as you did in the previous lesson, here you will be factoring the special case polynomials. Factoring is really the reverse of multiplication!
You can use algebra tiles or the box method. Recall that algebra tiles are very visual but can be cumbersome when you need to use a lot of tiles to find the factors.
Watch This
Khan Academy Factoring the Sum and Difference of Squares
Guidance
A box is to be designed for packaging with a side length represented by the quadratic \begin{align*}9b^2 - 64\end{align*}. If this is the most economical box, what are the dimensions?
First: factor the quadratic to find the value for \begin{align*}b\end{align*}.
\begin{align*}9b^2-64\end{align*}
First, the box:
The product of 3 and -64 is -576.
Need: factors for -576 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 576 since you need to have two numbers that are the same. This way they will add to give you 0 and multiply to give 576.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(3b+8)(3b-8)\end{align*}
So
\begin{align*}& 9b^2-64=(3b+8)(3b-8)\\ & \qquad \qquad \quad \swarrow \qquad \qquad \searrow\\ & \ \quad 3b+8=0 \qquad \qquad 3b-8=0\\ & \ \qquad \ \ 3b=-8 \qquad \qquad \quad 3b=8\\ & \ \qquad \quad b=\frac{-8}{3} \qquad \qquad \quad \ b=\frac{8}{3}\end{align*}
The most economical box is a cube. Therefore the dimensions are \begin{align*}\frac{8}{3} \times \frac{8}{3} \times \frac{8}{3}\end{align*}
Example A
Factor
\begin{align*}2x^2+28x+98\end{align*}
First, the box:
The product of 2 and 98 is 196.
Need: factors for 196 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 28. Start with taking the square root of 196 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 196. You could also find the factors of 196. This would give show you to two numbers that add to give you 28 and multiply to give you 196.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(2x+14)(2x+14)\end{align*}
So \begin{align*}2x^2+28x+98=(2x+14)(2x+14)\end{align*} Or
\begin{align*}2x^2+28x+98=(2x+14)^2\end{align*} SPECIAL CASE 1
Example B
Factor:
\begin{align*}8a^2-24a+18\end{align*}
First, let's factor out a 2 from this expression (the GCF!)
\begin{align*}8a^2-24a+18=2(4a^2-12a+9)\end{align*}
Next the box:
The product of 4 and 9 is 36.
Need: factors for 36 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -12. Start with taking the square root of 36 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 36. You could also find the factors of 36. This would give show you to two numbers that add to give you -12 and multiply to give you 36.
Next: GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}8a^2-24a+18=2(2a-3)(2a-3)\end{align*}
Or
\begin{align*}8a^2-24a+18=2(2a-3)^2\end{align*} SPECIAL CASE 2
Example C
Factor:
\begin{align*}x^2-16\end{align*}
First the box:
The product of 1 and 16 is 16.
Need: factors for 16 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 16 since you need to have two numbers that are the same. This way the middle terms could equal and cancel and the numbers would still multiply to give 16.
Next: GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}x^2-16=(x-4)(x+4)\end{align*} SPECIAL CASE 3
Vocabulary
- Difference of Two Squares
- The difference of two squares is a pattern found in polynomial expressions. It is a special case where there is no middle term found in the quadratic expression. The general equation for the difference of two squares is:
\begin{align*}x^2+y^2=(x+y)(x-y)\end{align*}
- Sum of Two Squares
- The sum of two squares involves two special patterns found in polynomial expressions. These are special cases where the middle term is twice the product of the first and last term. For the quadratic expression \begin{align*}ax^2+bx+c\end{align*}, the sum of two squares has a middle term \begin{align*}(b)\end{align*} equal to \begin{align*}2ac\end{align*}. These trinomials are often called perfect square trinomials. The general equations for the sum of two square patterns are:
- \begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}
- \begin{align*}(x-y)^2=x^2-2xy+y^2\end{align*}
Guided Practice
- Factor completely \begin{align*}s^2-18s+81\end{align*}
- Factor completely \begin{align*}50-98x^2\end{align*}
- Factor completely \begin{align*}4x^2+36x+144\end{align*}
Answers
1. \begin{align*}s^2-18s+81\end{align*}
First the box:
The product of 1 and 81 is 81.
Need: factors for 81 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -18. Start with taking the square root of 81 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 81. You could also find the factors of 81.
Next: GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}s^2-18s+81=(s-9)(s-9)\end{align*}
Or
\begin{align*}s^2-18s+81=(s-9)^2\end{align*} SPECIAL CASE 2
2. \begin{align*}50-98x^2\end{align*}
First: let's factor out a 2 from this expression (the GCF!)
\begin{align*}50-98x^2=2(25-49x^2)\end{align*}
Next the box:
The product of 25 and 49 is 1225. WOW!
Need: factors for 1225 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 1225 since you need to have two numbers that are the same. This way the middle terms could equal and cancel and the numbers would still multiply to give 16.
Next: GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}50-98x^2=2(5x-7)(5x+7)\end{align*} SPECIAL CASE 3
3. \begin{align*}4x^2+24x+44\end{align*}
First, let's factor out a 4 from this expression (the GCF!)
\begin{align*}4x^2+36x+144=4(x^2+12x+36)\end{align*}
Next the box:
The product of 1 and 36 is 36.
Need: factors for 36 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 6. Start with taking the square root of 36 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 36.
Next: GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(1x+6)(1x+6)\end{align*}
So \begin{align*}4x^2+36x+144=4(x+6)(x+6)\end{align*}
Or
\begin{align*}4x^2+36x+144=4(x+6)^2\end{align*} SPECIAL CASE 1
Summary
In this final lesson you have had the opportunity to apply your skills of factoring to the special cases that you learned about in the previous lesson. Remember the three special cases for polynomials:
A: The Sum of two Squares
1. \begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}
2. \begin{align*}(x-y)^2=x^2-2xy+y^2\end{align*}
B: The Difference of Two Squares
3. \begin{align*}x^2 + y^2 = (x + y)(x - y)x^2+y^2=(x+y)(x-y)\end{align*}
It is easier if you are able to recognize the special cases in that you do not need to find all of the factors for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} but simply need to find the square root of the “\begin{align*}ac\end{align*}” term. In special cases 1 and 2, the middle term is always 2“\begin{align*}ac\end{align*}”. For special case 3, there is no middle term. Both algebra tiles and the box method provide you with useful methods for factoring these special cases as they were with other polynomials.
Problem Set
Factor the following: (Special Case 1 and 2)
- \begin{align*}s^2+18s+81\end{align*}
- \begin{align*}x^2+12x+36\end{align*}
- \begin{align*}y^2-14y+49\end{align*}
- \begin{align*}4a^2+20a+25\end{align*}
- \begin{align*}9s^2-48s+64\end{align*}
Factor the following: (Special Case 3)
- \begin{align*}s^2-81\end{align*}
- \begin{align*}x^2-49\end{align*}
- \begin{align*}4t^2-25\end{align*}
- \begin{align*}25w^2-36\end{align*}
- \begin{align*}64-81a^2\end{align*}
Factor the following and identify the case:
- \begin{align*}y^2-22y+121\end{align*}
- \begin{align*}16t^2-49\end{align*}
- \begin{align*}9a^2+30a+25\end{align*}
- \begin{align*}100-25b^2\end{align*}
- \begin{align*}4s^2-28s+49\end{align*}
Answers
Factor...
- \begin{align*}s^2+18s+81\end{align*}
Need the box:
The product of 1 and 81 is 81.
Need: factors for 81 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 18. Start with taking the square root of 81 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 81.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(1x+9)(1x+9)\end{align*}
So \begin{align*}s^2+18s+81=(s+9)(s+9)\end{align*}
Or
\begin{align*}s^2+18s+81=(s+9)^2\end{align*} SPECIAL CASE 1
- \begin{align*}y^2-14y+49\end{align*}
Need the box:
The product of 1 and 49 is 49.
Need: factors for 49 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -14. Start with taking the square root of 49 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 49.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(1y-7)(1y-7)\end{align*}
So \begin{align*}y^2-14y+49=(y-7)(y-7)\end{align*}
Or
\begin{align*}y^2-14y+49=(y-7)^2\end{align*} SPECIAL CASE 2
- \begin{align*}9s^2-48s+64\end{align*}
Need the box:
The product of 9 and 64 is 576.
Need: factors for 576 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -48. Start with taking the square root of 576 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 576.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(3s-8)(3s-8)\end{align*}
So \begin{align*}9s^2-48s+64=(3s-8)(3s-8)\end{align*}
Or
\begin{align*}9s^2-48s+64=(3s-8)^2\end{align*} SPECIAL CASE 2
Factor...
- \begin{align*}s^2-81\end{align*}
Need the box:
The product of 1 and 81 is 81.
Need: factors for 81 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 81 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 81.
Next: the GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}s^2-81=(s-9)(s+9)\end{align*}
Or \begin{align*}s^2-81=(s-9)(s+9)\end{align*} SPECIAL CASE 3
- \begin{align*}4t^2-25\end{align*}
Need the box:
The product of 4 and 25 is 100.
Need: factors for 100 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 100 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 100.
Next: the GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}4t^2-25=(2t-5)(2t+5)\end{align*}
Or
\begin{align*}4t^2-25=(2t-5)(2t+5)\end{align*} SPECIAL CASE 3
- \begin{align*}64-81a^2\end{align*}
Need the box:
The product of 81 and 64 is 5184. WOW!
Need: factors for 5184 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 0. Start with taking the square root of 5184 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 5184.
Next: the GCFs
Last: Put it together
These new numbers represent your factors.
So \begin{align*}64-81a^2=(8-9a)(8+9a)\end{align*}
Or
\begin{align*}64-81a^2=(8-9a)(8+9a)\end{align*} SPECIAL CASE 3
Factor...
- \begin{align*}y^2-22y+121\end{align*}
Need the box:
The product of 1 and 121 is 121.
Need: factors for 121 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of −22. Start with taking the square root of 121 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 121.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(1y-11)(1y-11)\end{align*}
So \begin{align*}y^2-22y+121=(y-11)(y-11)\end{align*}
Or
\begin{align*}y^2-22y+121=(y-11)^2\end{align*} SPECIAL CASE 2
- \begin{align*}9a^2+30a+25\end{align*}
Need the box:
The product of 9 and 25 is 225.
Need: factors for 225 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of 30. Start with taking the square root of 225 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 225.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(3a+5)(3a+5)\end{align*}
So \begin{align*}9a^2+30a+25=(3a+5)(3a+5)\end{align*}
Or
\begin{align*}9a^2+30a+25=(3a+5)^2\end{align*} SPECIAL CASE 1
- \begin{align*}4s^2-28s+49\end{align*}
Need the box:
The product of 4 and 49 is 196.
Need: factors for 196 and a pair that will combine to give the ‘\begin{align*}b\end{align*}’ value of -28. Start with taking the square root of 196 since you need to have two numbers that are the same. This way the middle terms would be the same and would still multiply to give 196.
Next: the GCFs
Last: Put it together
These new numbers represent your factors. \begin{align*}(2s-7)(2s-7)\end{align*}
So \begin{align*}4s^2-28s+49=(2s-7)(2s-7)\end{align*}
Or
\begin{align*}4s^2-28s+49=(3s-8)^2\end{align*} SPECIAL CASE 2
Summary
In this lesson, you have worked with adding and subtracting polynomials. You also worked with multiplying polynomials. Remember that when multiplying polynomials, the distributive property is the tool to use. You began your look at the special cases of polynomials and learned that the more you recognize these special cases, the quicker it is to work with them when you are factoring.
In this current lesson, you continued your learning of polynomials by working with the greatest common factor. Make the question simpler by first removing the number or letter that is common in all terms of your polynomial. You also began factoring quadratics (or trinomials) where the coefficient for \begin{align*}a\end{align*} in \begin{align*}ax^2 + bx + c\end{align*} was equal to 1 and then for cases where a was not equal to 1. Quadratics where \begin{align*}a\end{align*} was not equal to one seemed a bit more complex but remember the strategy for solving these remains the same.
Lastly, in this current lesson, you worked with factoring polynomials for three specis cases. You will have noticed that these special cases were those you worked with in lesson Addition and Subtraction of Polynomials, only here you were factoring them. Again, strategies for factoring the special case polynomials remain the same. As well, if you are able to recognize a quadratic (trinomial) as a special case, factoring these is somewhat less cumbersome.
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