7.5: Factorization of Quadratic Expressions
Jack wants to construct a border around two sides of his garden. The garden measures 5 yards by 18 yards. He has enough stone to build a border with a total area of 30 square yards. The border will be twice as wide on the shorter end. What are the dimensions of the border?
Watch This
Khan Academy Factoring trinomials with a leading 1 coefficient
James Sousa: Factoring Trinomials using Trial and Error and Grouping
Guidance
To factor a polynomial means to write the polynomial as a product of other polynomials. Here, you'll focus on factoring quadratic expressions. Quadratic expressions are polynomials of degree 2, of the form \begin{align*}ax^2+bx+c\end{align*}
\begin{align*}(2x+3)(3x-5)&=6x^2-10x+9x-15\\ &=6x^2-x-15\end{align*}
When factoring a quadratic expression, your job will be to take an expression like \begin{align*}6x^2-x-15\end{align*} and write it as \begin{align*}(2x+3)(3x-5)\end{align*}. You can think of factoring as the reverse of multiplying. Notice that when factored, the \begin{align*}6x^2\end{align*} factors to \begin{align*}2x\end{align*} and \begin{align*}3x\end{align*}. The \begin{align*}-15\end{align*} factors to \begin{align*}-5\end{align*} and \begin{align*}3\end{align*}. You can say then, in general, 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 \begin{align*}b = {\color{red}d} {\color{blue}g} + {\color{blue}e} {\color{red}f}\end{align*}
Here you will work through a number of examples to develop mastery at factoring trinomials using a box method.
Example A
Factor: \begin{align*}2x^2+11x+15\end{align*}
Solution: First note that there is not a common factor in this trinomial. If there was, you would want to start by factoring out the common factor. In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 2 and the ‘\begin{align*}c\end{align*}’ value is 15. Start by making a box and placing these values in the box as shown.
The product of 2 and 15 is 30. To continue filling in the box, you need to find two numbers that multiply to 30, but add up to +11 (the value of \begin{align*}b\end{align*} in the original equation). The two numbers that work are 5 and 6: \begin{align*}5+6=11\end{align*} and \begin{align*}5\cdot 6=30\end{align*}. Put 5 and 6 in the box.
Next, find the GCF of the numbers in each row and each column and put these new numbers in the box. The first row, 2 and 6, has a GCF of 2. The second row, 5 and 15, has a GCF of 5.
The first column, 2 and 5, has a GCF of 1. The second column, 6 and 15, has a GCF of 3.
Notice that the products of each row GCF with each column GCF create the original 4 numbers in the box. The GCFs represent the coefficients of your factors. Your factors are \begin{align*}(1x + 3)\end{align*} and \begin{align*}(2x + 5)\end{align*}. You can verify that those binomials multiply to create the original trinomial: \begin{align*}(x+3)(2x+5)=2x^2+5x+6x+15=2x^2+11x+15\end{align*}.
The factored form of \begin{align*}2x^2+11x+15\end{align*} is \begin{align*}(x+3)(2x+5)\end{align*}.
Example B
Factor: \begin{align*}3x^2-8x-3\end{align*}
Solution: First note that there is not a common factor in this trinomial. If there was, you would want to start by factoring out the common factor. In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 3 and the ‘\begin{align*}c\end{align*}’ value is –3. Start by making a box and placing these values in the box as shown.
The product of 3 and –3 is –9. To continue filling in the box, you need to find two numbers that multiply to –9, but add up to –8 (the value of \begin{align*}b\end{align*} in the original equation). The two numbers that work are –9 and 1. \begin{align*}-9+1=-8\end{align*} and \begin{align*}-9\cdot 1=-9\end{align*}. Put –9 and 1 in the box.
Next, find the GCF of the numbers in each row and each column and put these new numbers in the box. The first row, 3 and 1, has a GCF of 1. The second row, –9 and –3, has a GCF of –3.
The first column, 3 and –9, has a GCF of 3. The second column, 1 and –3, has a GCF of 1.
Notice that the products of each row GCF with each column GCF create the original 4 numbers in the box. The GCFs represent the coefficients of your factors. Your factors are \begin{align*}(3x + 1)\end{align*} and \begin{align*}(1x-3)\end{align*}. You can verify that those binomials multiply to create the original trinomial: \begin{align*}(3x+1)(x-3)=3x^2-9x+1x-3=3x^2-8x-3\end{align*}.
The factored form of \begin{align*}3x^2-8x-3\end{align*} is \begin{align*}(3x+1)(x-3)\end{align*}.
Example C
Factor: \begin{align*}5w^2-21w+18\end{align*}
Solution: First note that there is not a common factor in this trinomial. If there was, you would want to start by factoring out the common factor. In this trinomial, the ‘\begin{align*}a\end{align*}’ value is 5 and the ‘\begin{align*}c\end{align*}’ value is 18. Start by making a box and placing these values in the box as shown.
The product of 5 and 18 is 90. To continue filling in the box, you need to find two numbers that multiply to 90, but add up to –21 (the value of \begin{align*}b\end{align*} is the original equation). The two numbers that work are –6 and –15. \begin{align*}-6+(-15)=-21\end{align*} and \begin{align*}-6\cdot -15=90\end{align*}. Put –6 and –15 in the box.
Next, find the GCF of the numbers in each row and each column and put these new numbers in the box. The first row, 5 and –6, has a GCF of 1. The second row, –15 and 18, has a GCF of 3.
The first column, 5 and –15, has a GCF of 5. The second column, –6 and 18, has a GCF of 6.
Notice that you need to make two of the GCFs negative in order to make the products of each row GCF with each column GCF create the original 4 numbers in the box. The GCFs represent the coefficients of your factors. Your factors are \begin{align*}(5w-6)\end{align*} and \begin{align*}(w-3)\end{align*}. You can verify that those binomials multiply to create the original trinomial: \begin{align*}(5w-6)(w-3)=5w^2-15w-6w+18=5w^2-21w+18\end{align*}.
The factored form of \begin{align*}5w^2-21w+18\end{align*} is \begin{align*}(5w-6)(w-3)\end{align*}.
Concept Problem Revisited
Jack wants to construct a border around two sides of his garden. The garden measures 5 yards by 18 yards. He has enough stone to build a border with a total area of 30 square yards. The border 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} &= 30 \ yd^2\\ \text{Area of Garden} + \text{border} &= (18 + 2x)(5 + x)\\ \text{Area of border} &= (\text{Area of garden} + \text{border}) - \text{Area of garden}\\ 30 &= (18 + 2x)(5 + x) - 90\\ 30 &= 90+18x+10x+2x^2-90\\ 30 &= 28x+2x^2\\ 0 &= 2x^2+28x-30\end{align*}
This trinomial has a common factor of 2. First, factor out this common factor:
\begin{align*}2x^2+28x-30=2(x^2+14x-15)\end{align*}
Now, you can use the box method to factor the remaining trinomial. After using the box method, your result should be:
\begin{align*}2(x^2+14x-15)=2(x+15)(x-1)\end{align*}
To find the dimensions of the border you need to solve a quadratic equation. This is explored in further detail in another concept:
\begin{align*}& \ 2(x+15)(x-1)=0\\ & \ \swarrow \qquad \qquad \searrow\\ & x+15=0 \ \ x-1=0\\ & x=-15 \quad \ \ x=1\end{align*}
Since \begin{align*}x\end{align*} cannot be negative, \begin{align*}x\end{align*} must equal \begin{align*}1\end{align*}.
Width of Border: \begin{align*}2x = 2(1) = 2 \ yd\end{align*}
Length of Border: \begin{align*}x = 1 \ yd\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 polynomial of degree 2. The general form of a quadratic expression is \begin{align*}ax^2 + bx + c\end{align*}.
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:
1. Use the box method and you find that \begin{align*}8c^2-2c-3=(2c+1)(4c-3)\end{align*}
2. First you can factor out the 3 from the polynomial. Then, use the box method. The final answer is \begin{align*}3m^2+3m-60=3(m-4)(m+5)\end{align*}.
3. First you can factor out the \begin{align*}5e\end{align*} from the polynomial. Then, use the box method. The final answer is \begin{align*}5e^3+30e^2+40e=5e(e+2)(e+4)\end{align*}.
Practice
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*}
- \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*}
- \begin{align*}x^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*}
- \begin{align*}3x^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*}
- \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*}
- \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*}
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