7.6: Cubic Polynomials
Introduction
In this lesson you will learn to factor cubic polynomials by removing a common factor. You will also learn to factor these polynomials by grouping. The final concept that you will learn will be to apply factoring by grouping to special products.
Objectives
The lesson objectives for Factoring Polynomials are:
- Factoring by Removing a Common Factor
- Factoring by Grouping
- Factoring Special Products
Factoring a Common Factor
Introduction
In this concept you will be working with polynomials of the third degree. Polynomials where the largest exponent on the variable is three (3) are known as cubics. Therefore a cubic polynomial is a polynomial of degree equal to 3. An example could include \begin{align*}9x^3+10x-5\end{align*}. Another example is \begin{align*}8x^3+2x^2-5x-7\end{align*}. In this first concept of lesson Cubic Polynomials, you will be factoring cubics by removing a common factor. Recall in lesson Factoring Polynomials when you removed common factors from quadratic (trinomial) functions. The same skills will be applied here. You use the distributive property to factor out the greatest common factor (GCF) when factoring cubics that have common terms in the polynomials.
Watch This
Khan Academy Factoring and the Distributive Property
Guidance
Factor completely the following polynomial: \begin{align*}8x^3+24x^2-32x\end{align*}.
Look for the common factors in each of the terms.
\begin{align*}8x^3 &= {\color{red}8} \cdot x \cdot x \cdot {\color{red}x} \\ 24x^2 &= {\color{red}8} \cdot 3 \cdot x \cdot {\color{red}x} \\ -32x &= {\color{red}8} \cdot -4 \cdot {\color{red}x}\end{align*}
Therefore:
\begin{align*}8x^3 +24x^2 + 32x &= 8x(x^2+3x-4) \\ &= 8x(x+4)(x-1)\end{align*}
Example A
Factor completely the following polynomial: \begin{align*}3x^3-15x\end{align*}.
Look for the common factors in each of the terms.
\begin{align*}3x^3 &= {\color{red}3} \cdot x \cdot x \cdot {\color{red}x} \\ -15x &= {\color{red}3} \cdot -5 \cdot {\color{red}x}\end{align*}
Therefore:
\begin{align*}3x^3-15x=3x(x^2-5)\end{align*}
Example B
Factor completely the following polynomial: \begin{align*}2a^3+16a^2+8a\end{align*}.
Look for the common factors in each of the terms.
\begin{align*}2a^3 &= {\color{red}2} \cdot a \cdot a \cdot {\color{red}a} \\ 16a^2 &= {\color{red}2} \cdot 8 \cdot a \cdot {\color{red}a} \\ 8a &= {\color{red}2} \cdot 4 \cdot {\color{red}a}\end{align*}
Therefore:
\begin{align*}2a^3+16a^2+8a=2a(a^2+8a+4)\end{align*}
Example C
Factor completely the following polynomial: \begin{align*}6s^3+36s^2-18s-42\end{align*}.
Look for the common factors in each of the terms.
\begin{align*}6s^3 &= {\color{red}6} \cdot s \cdot s \cdot s \\ 36s^2 &= {\color{red}6} \cdot 6 \cdot s \cdot s \\ -18s &= {\color{red}6} \cdot -3 \cdot s \\ -42 &= {\color{red}6} \cdot -7\end{align*}
Therefore:
\begin{align*}6s^3+36s^2-18s-42=6(s^3+6s^2-3s-7)\end{align*}.
Vocabulary
- Cubic Polynomial
- A cubic polynomial is a polynomial of degree equal to 3. For example \begin{align*}8x^3+2x^2-5x-7\end{align*} is a cubic polynomial.
- Distributive Property
- The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*} {\color{red}\frac{2}{3}}({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
- 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
1. Factor completely the following polynomial: \begin{align*}9w^3+12w\end{align*}.
2. Factor completely the following polynomial: \begin{align*}y^3+4y^2+4y\end{align*}.
3. Factor completely the following polynomial: \begin{align*}2t^3-10t^2+8t\end{align*}.
Answers
1. \begin{align*}9w^3+12w\end{align*}
Look for the common factors in each of the terms.
\begin{align*}9w^3 &= {\color{red}3} \cdot 3 \cdot w \cdot w \cdot {\color{red}w} \\ 12w &= {\color{red}3} \cdot 4 \cdot {\color{red}w}\end{align*}
Therefore:
\begin{align*}9w^3+12w=3w(3w^2+4)\end{align*}
2. \begin{align*}y^3+4y^2+4y\end{align*}
Look for the common factors in each of the terms.
\begin{align*}y^3 &= y \cdot y \cdot {\color{red}y} \\ 4y^2 &= 4 \cdot y \cdot {\color{red}y} \\ 4y &= 4 \cdot {\color{red}y}\end{align*}
Therefore:
\begin{align*}y^3+4y^2+4y=y(y^2+4y+4)\end{align*}
3. \begin{align*}2t^3-10t^2+8t\end{align*}
Look for the common factors in each of the terms.
\begin{align*}2t^3 &= {\color{red}2} \cdot t \cdot t \cdot {\color{red}t} \\ -10t^2 &= {\color{red}2} \cdot -5 \cdot t \cdot {\color{red}t} \\ 8t &= {\color{red}2} \cdot 4 \cdot {\color{red}t}\end{align*}
Therefore:
\begin{align*}2t^3 -10t^2 +8t &= 2t(t^2-5t+4) \\ &= 2t(t-4)(t-1)\end{align*}
Summary
In this concept you may have noticed that you used similar problem solving strategies to those you used in lesson Factoring Polynomials. You began your study of cubic polynomials by factoring out the greatest common factor (the GCF). This strategy you learned earlier.
One of the easiest ways to visualize the greatest common factor is to isolate each term and break each term down into its smallest parts. In this way, you can quickly see which factors are common to all terms in the polynomial and therefore factor them out of the cubic.
Problem Set
Factor completely each of the following polynomials:
- \begin{align*}6x^3-12\end{align*}
- \begin{align*}4x^3-12x^2\end{align*}
- \begin{align*}8y^3+32y\end{align*}
- \begin{align*}15a^3+30a^2\end{align*}
- \begin{align*}21q^3+63q\end{align*}
Factor completely each of the following polynomials:
- \begin{align*}4x^3-12x^2-8\end{align*}
- \begin{align*}12e^3+24e^2-6\end{align*}
- \begin{align*}15s^3-30s+45\end{align*}
- \begin{align*}22r^3-11r^2+44\end{align*}
- \begin{align*}32d^3-16d^2+12d\end{align*}
Factor completely each of the following polynomials:
- \begin{align*}5x^3+15x^2+25x-30\end{align*}
- \begin{align*}3y^3-9y^2+27y+36\end{align*}
- \begin{align*}12s^3-24s^2+36s-48\end{align*}
- \begin{align*}8x^3+24x^2+80x\end{align*}
- \begin{align*}5x^3-25x^2-70x\end{align*}
Factoring by Grouping
Introduction
Another way that you can use the distributive property to factor polynomials is to factor by grouping. Factoring by grouping is used when you have polynomials with four terms. It cannot be used all of the time but when it can, finding the greatest common factor is possible. In factoring by grouping you factor the terms into two groups and then take the greatest common factor out of one of them.
In this concept you will learn to work with finding the greatest common factor of cubic polynomials using the concept of factoring by grouping. In factoring by grouping, which is different from what you have just learned, you have to first group the terms into two groups. You then factor each group, use your distributive property, then factor any quadratics that remain (and are possible). Let’s begin exploring factoring by grouping cubic polynomials!
Watch This
Khan Academy Factoring by Grouping
Note: The above video shows factoring by grouping of quadratic (trinomial) expressions. The same problem solving concept will be developed in this lesson for cubic polynomials.
Guidance
A tank is bought at the pet store and is known to have a volume of 12 cubic feet. The dimensions are shown in the diagram below. If your new pet requires the tank to be at least 3 feet high, did you buy a big enough tank?
To solve this problem, you need to calculate the volume of the tank.
\begin{align*}V &= l\times w\times h \\ 12 &= (x+4) \times (x-1) \times (x) \\ 12 &= (x^2 +3x -4) \times (x) \\ 12 &= x^3 + 3x^2 -4x \\ 0 &= x^3 + 3x^2 -4x -12\end{align*}
Now you need to factor by grouping.
\begin{align*}0 = (x^3 + 3x^2) - (4x + 12)\end{align*}
Factor out the common terms in each of the sets of brackets.
\begin{align*}0 = x^2 (x+3) - 4(x + 3)\end{align*}
Factor out the group of terms \begin{align*}(x + 3)\end{align*} from the expression.
\begin{align*}0 = (x+3)(x^2-4)\end{align*}
Completely factor the remaining quadratic expression.
\begin{align*}0 = (x+3)(x-2)(x+2)\end{align*}
Now solve for the variable \begin{align*}x\end{align*}.
\begin{align*}& \qquad 0 = (x+3) \ (x-2) \ (x+2) \\ & \qquad \qquad \swarrow \qquad \quad \ \downarrow \qquad \quad \ \searrow \\ & \quad \ \ x+3=0 \quad x-2=0 \quad x+2=0 \\ & \qquad \quad \ x=-3 \qquad x=2 \qquad \ \ x=-2 \end{align*}
Since you are looking for a length, only \begin{align*}x = 2\end{align*} is a good solution. But since you need a tank 3 feet high and this one is only 2 feet high, you need to go back to the pet shop and buy a bigger one.
Example A
Factor the following polynomial by grouping:
\begin{align*}w^3-2w^2-9w+18\end{align*}.
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
\begin{align*}w^3-2w^2-9w+18=(w^3-2w^2)-(9w-18)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(w^3-2w^2)-(9w-18) = w^2(w-2)-9(w-2)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(w - 2)\end{align*}.
\begin{align*}w^2(w-2)-9(w-2) = (w-2)(w^2-9)\end{align*}
Step 4: Completely factor the remaining quadratic expression \begin{align*}(w^2-9)\end{align*}.
\begin{align*}(w-2)(w^2-9) = (w-2)(w+3)(w-3)\end{align*}
Therefore:
\begin{align*}w^3-2w^2-9w+18 = (w-2)(w+3)(w-3)\end{align*}
Example B
Factor the following polynomial by grouping:
\begin{align*}2s^3-8s^2+3s-12\end{align*}
Step 1: Group the terms into two groups.
\begin{align*}2s^3-8s^2+3s-12 = (2s^3-8s^2)+(3s-12)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(2s^3-8s^2) + (3s-12) = 2s^2(s-4)+3(s-4)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(s-4)\end{align*}.
\begin{align*}2s^2(s-4) + 3(s-4) = (s-4) (2s^2+3)\end{align*}
Step 4: The expression \begin{align*}(2s^3+3)\end{align*} cannot be factored.
Therefore:
\begin{align*}2s^3-8s^2+3s-12 = (s-4)(2s^2+3)\end{align*}
Example C
Factor the following polynomial by grouping: \begin{align*}y^3+5y^2-4y-20\end{align*}.
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
\begin{align*}y^3+5y^2-4y-20=(y^3+5y^2)-(4y+20)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(y^3+5y^2)-(4y+20) = y^2(y+5)-4(y+5)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(y + 5)\end{align*}.
\begin{align*}y^2(y+5)-4(y+5)=(y+5)(y^2-4)\end{align*}
Step 4: Completely factor the remaining quadratic expression \begin{align*}(y^2-4)\end{align*}.
\begin{align*}(y+5)(y^2-4)=(y+5)(y+2)(y-2)\end{align*}
Therefore:
\begin{align*}y^3+5y^2-4y-20=(y+5)(y+y)(y-2)\end{align*}
Vocabulary
- Distributive Property
- The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*} {\color{red}\frac{2}{3}}({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}\left({\color{red}\frac{2}{3}}\right)\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
- 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
1. Factor the following polynomial by grouping: \begin{align*}y^3-4y^2-4y+16\end{align*}.
2. Factor the following polynomial by grouping: \begin{align*}3x^3-4x^2-3x+4\end{align*}.
3. Factor the following polynomial by grouping: \begin{align*}e^3+3e-4e-12\end{align*}.
Answers
1. \begin{align*}y^3-4y^2-4y+16\end{align*}
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
\begin{align*}y^3-4y^2-4y+12=(y^3-4y^2)-(4y-16)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(y^3-4y^2)-(4y-12)=y^2(y-4)-4(y-4)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(y - 4)\end{align*}.
\begin{align*}y^2(y-4)-4(y-4)=(y-4)(y^2-4)\end{align*}
Step 4: Completely factor the remaining quadratic expression \begin{align*}(y^2-4)\end{align*}.
\begin{align*}(y-4)(y^2-4)=(y-4)(y+2)(y-2)\end{align*}
Therefore:
\begin{align*}y^3-4y^2-4y+16 = (y-4)(y+2)(y-2)\end{align*}
2. \begin{align*}3x^3-4x^2-3x+4\end{align*}
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
\begin{align*}3x^3-4x^2-3x+4=(3x^3-4x^2)-(3x-4)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(3x^3-4x^2)-(3x-4)=x^2(3x-4)-1(3x-4)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(3x-4)\end{align*}.
\begin{align*}x^2(3x-4)-1(3x-4)=(3x-4)(x^2-1)\end{align*}
Step 4: Completely factor the remaining quadratic expression \begin{align*}(x^2-1)\end{align*}.
\begin{align*}(3x-4)(x^2-1)=(3x-4)(x+1)(x-1)\end{align*}
Therefore:
\begin{align*}3x^3-4x^2-3x+4=(3x-4)(x+1)(x-1)\end{align*}
3. \begin{align*}e^3+3e-4e-12\end{align*}
Step 1: Group the terms into two groups. Notice the sign change on the second group because of the negative sign.
\begin{align*}e^3+3e^2-4e-12 = (e^3+3e^2) - (4e+12)\end{align*}
Step 2: Factor out the common terms in each of the sets of brackets.
\begin{align*}(e^3+3e^2)-(4e+12) = e^2(e+3) -4(e+3)\end{align*}
Step 3: Use the distributive property to factor out the common group \begin{align*}(e + 3)\end{align*}.
\begin{align*}e^2(e+3)-4(e+3) = (e+3)(e^2-4)\end{align*}
Step 4: Completely factor the remaining quadratic expression \begin{align*}(e^2-4)\end{align*}.
\begin{align*}(e+3)(e^2-4)=(e+3)(e+2)(e-2)\end{align*}
Therefore:
\begin{align*}e^3+3e^2-4e-12=(e+3)(e+2)(e-2)\end{align*}
Summary
To factor cubic polynomials by grouping involves four steps, one of which is the distributive property. The distributive property is something you have been learning for a long time in Algebra, and its application in cubic polynomials is just one more way it shows its usefulness. Factoring cubic polynomials involves problem solving skills that you have learned in previous lessons such as factoring quadratics, finding greatest common factors, and combining like terms. For these problems, however, you have the opportunity to combine all of these skills in one problem solving experience!
To solve cubic polynomials by grouping remember the four steps:
Step 1: Group the terms into two groups. Remember to watch for the sign change when you are grouping the second set of terms. Placing the negative outside a bracket will change the signs inside the bracket.
Step 2: Factor out the common terms in each of the sets of brackets.
Step 3: Use the distributive property to factor out the common group (the binomial).
Step 4: Completely factor the remaining quadratic expression.
Problem Set
Factor the following cubic polynomials by grouping:
- \begin{align*}x^3-3x^2-36x+108\end{align*}
- \begin{align*}e^3-3e^2-81e+243\end{align*}
- \begin{align*}x^3-10x^2-49x+490\end{align*}
- \begin{align*}y^3-7y^2-5y+35\end{align*}
- \begin{align*}x^3+9x^2+3x+27\end{align*}
Factor the following cubic polynomials by grouping:
- \begin{align*}3x^3+x^2-3x-1\end{align*}
- \begin{align*}5s^3-6s^2-45s+54\end{align*}
- \begin{align*}4a^3-7a^2+4a-7\end{align*}
- \begin{align*}5y^2+15y^2-45y-135\end{align*}
- \begin{align*}3x^3+15x^2-12x-60\end{align*}
Factor the following cubic polynomials by grouping:
- \begin{align*}2e^3+14e^2+7e+49\end{align*}
- \begin{align*}2k^3+16k^2+38k+24\end{align*}
- \begin{align*}-6x^3+3x^2+54x-27\end{align*}
- \begin{align*}-5m^3-6m^2+20m+24\end{align*}
- \begin{align*}-2x^3-8x^2+14x+56\end{align*}
Factoring Special Products
Introduction
Just as you learned the special products for quadratic (trinomial) expressions, there are special products for cubic polynomials. Remember there were three special products for quadratic (trinomial) expressions. For cubic polynomials, there are two special products. The first is the sum of two cubes. The second is the difference of two cubes. These two special products are shown in the box below.
The sum of two cubes
\begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}
The difference of two cubes
\begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}
In this concept you will learn to factor the two types of special products in cubic polynomials. As it was with the quadratic expressions, being able to recognize that the cubic polynomial is indeed a special product means that you can quickly factor the polynomial. Take a moment to commit these two special cases to memory so that solving these types of cubic polynomials will be more efficient.
Guidance
Factor the following cubic polynomial: \begin{align*}375x^3+648\end{align*}.
To solve this problem you need to first recognize that this is a special case cubic polynomial, namely \begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
Next if you look at this polynomial, you see that first you can factor out the greatest common factor.
\begin{align*}& 375x^3+648: \\ & 375x^3 = {\color{red}3} \cdot 125x^3 \\ & 648 = {\color{red}3} \cdot 216\end{align*}
Therefore \begin{align*}375x^3+648=3(125x^3+216)\end{align*}
Now let’s factor the special case polynomial. Remember that you have a number (numerical coefficient) attached to the \begin{align*}x\end{align*} variable. This means that you have to include it in the factoring. In other words for now becomes \begin{align*}(x+y)=(ax+y)\end{align*}, and \begin{align*}(x^2-xy+y^2)\end{align*} now becomes \begin{align*}(a^2x^2-axy+y^2)\end{align*}.
\begin{align*}375x^3 +648 = 3(125x^3+216)\end{align*}
\begin{align*}375x^3 +648 = 3(5x+6)(25x^2-30x+36)\end{align*}
Example A
\begin{align*}x^3+27\end{align*}
This is the sum of two cubes or \begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
Step 1: Put the cubic polynomial into the sum of two cubes form:
\begin{align*}x^3+27=x^3+y^3\end{align*}
Step 2: Factor out the sum of two cubes
\begin{align*}x^3+3^3=(x+3)(x^2-3x+9)\end{align*}.
Example B
\begin{align*}x^3-343\end{align*}
This is the difference of two cubes or \begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
Step 1: Put the cubic polynomial into the sum of two cubes form:
\begin{align*}x^3-343=x^3-7^3\end{align*}
Step 2: Factor out the sum of two cubes
\begin{align*}x^3-343=(x-7)(x^2+7x+49)\end{align*}.
Example C
\begin{align*}64x^3-1\end{align*}
This is the difference of two cubes or \begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
Step 1: Put the cubic polynomial into the sum of two cubes form:
\begin{align*}64x^3-1=4^3x^3-1^3\end{align*}
Step 2: Factor out the sum of two cubes
\begin{align*}64x^3-1=(4x-1)(4^2x^2+4x+1^2)\end{align*}.
\begin{align*}64x^3-1=(4x-1)(16x^2+4x+1)\end{align*}.
Vocabulary
- Distributive Property
- The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}3}({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}3})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}3})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
- Difference of Two Cubes
- The difference of two cubes is a special product cubic polynomial in the form of \begin{align*}x^3-y^3\end{align*}. This type of polynomial can be quickly factored using the expression:
\begin{align*}(x^3-y^3)=(x-y)(x^2+xy+y^2)\end{align*}
- Sum of Two Cubes
- The sum of two cubes is a special product cubic polynomial in the form of \begin{align*}x^3+y^3\end{align*}. This type of polynomial can be quickly factored using the expression:
\begin{align*}(x^3+y^3)=(x+y)(x^2-xy+y^2)\end{align*}
Guided Practice
1. Factor the following special product cubic polynomial: \begin{align*}x^3+512\end{align*}
2. Factor the following special product cubic polynomial: \begin{align*}8x^3+125\end{align*}
3. Factor the following special product cubic polynomial: \begin{align*}x^3-216\end{align*}
Answers
1. \begin{align*}x^3+512\end{align*}
This is the sum of two cubes or \begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}.
Step 1: Put the cubic polynomial into the sum of two cubes form: \begin{align*}x^3+512=x^3+8^3\end{align*}
Step 2: Factor out the sum of two cubes \begin{align*}x^3+8^3=(x+8)(x^2-8x+64)\end{align*}.
2. \begin{align*}8x^3+125\end{align*}
This is the sum of two cubes or \begin{align*}x^3+y^3=(x+y)(x^2-xy+y^2)\end{align*}
Step 1: Put the cubic polynomial into the sum of two cubes form: \begin{align*}8x^3+125=2^3x^3+5^3\end{align*}
Step 2: Factor out the sum of two cubes \begin{align*}2^3x^3+5^3=(2x+5)(4x^2-10x+25)\end{align*}.
3. \begin{align*}x^3-216\end{align*}
This is the difference of two cubes or \begin{align*}x^3-y^3=(x-y)(x^2+xy+y^2)\end{align*}.
Step 1: Put the cubic polynomial into the sum of two cubes form: \begin{align*}x^3-216=x^3-6^3\end{align*}
Step 2: Factor out the sum of two cubes \begin{align*}x^3-6^3=(x-6)(x^2+6x+36)\end{align*}.
Summary
In lesson Operations with Polynomials you learned about special cases in quadratic expressions or polynomials with a degree of two. In this final concept of lesson Cubic Polynomials, you learned about the special cases for polynomials with a degree of 3. The box below summarizes the special cases for both quadratic and cubic polynomials.
Special Case 1: \begin{align*}(x+y)^2=x^2+2xy+y^2\end{align*}
Example: \begin{align*}(x+5)^2=x^2+10x+25\end{align*}
Special Case 2: \begin{align*}(x-y)^2=x^2-2xy+y^2\end{align*}
Example: \begin{align*}(2x-8)^2=4x^2-32x+64\end{align*}
Special Case 3: \begin{align*}(x+y)(x-y)=x^2-y^2\end{align*}
Example: \begin{align*}(5x+10)(5x-10)=25x^2-100\end{align*}
Sum of two cubes: \begin{align*}(x^3+y^3)=(x+y)(x^2-xy+y^2)\end{align*}
Example: \begin{align*}(x^3+125)=(x+5)(x^2-5x+25)\end{align*}
Difference of two cubes: \begin{align*}(x^3-y^3)=(x-y)(x^2+xy+y^2)\end{align*}
Example: \begin{align*}(x^3-125)=(x-5)(x^2+5x+25)\end{align*}
Learning these special cases and being able to recognize them allows you to quickly factor such polynomials in problem solving situations.
Problem Set
Factor the following special product cubic polynomials:
- \begin{align*}x^3+h^3\end{align*}
- \begin{align*}a^3+125\end{align*}
- \begin{align*}8x^3+64\end{align*}
- \begin{align*}x^3+1728\end{align*}
- \begin{align*}2x^3+6750\end{align*}
Factor the following special product cubic polynomials:
- \begin{align*}h^3-64\end{align*}
- \begin{align*}s^3-216\end{align*}
- \begin{align*}p^3-512\end{align*}
- \begin{align*}4e^3-32\end{align*}
- \begin{align*}2w^3-250\end{align*}
Factor the following special product cubic polynomials:
- \begin{align*}x^3+8\end{align*}
- \begin{align*}y^3-1\end{align*}
- \begin{align*}125e^3-8\end{align*}
- \begin{align*}64a^3+2197\end{align*}
- \begin{align*}54z^3+3456\end{align*}
Summary
In this lesson, you have worked with cubic polynomials. Cubic polynomials are polynomials of degree three. Examples include \begin{align*}x^3+8, x^3-4x^2+3x-5\end{align*}, and so on. Notice in these examples, the largest exponent for the variable is three (3). They are all cubics. You worked with three concepts in this lesson. The first concept involved factoring a common factor. You worked on this problem solving strategy earlier with quadratic (trinomial) expressions, and the same procedure is involved with cubic polynomials.
The second concept required you to factor by grouping. This concept can be done with quadratics as well as cubics. It allows you to quickly factor a cubic by grouping the four terms into two groups and then factoring out a common binomial from each group.
The last concept introduced you to the two special products of cubic polynomials. You learned three special cases for quadratics and now you have two special cases for cubics. What is beneficial about these special cases is that they allow you to quickly factor the cubic polynomials by following the associated rule. Knowing how to identify polynomials in these special cases will allow you to quickly factor both quadratic and cubic polynomials.
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