CR.12: Factoring Higher Power Polynomials and Special Polynomials

Learning Outcomes

Factor a perfect square trinomial.
Factor a difference of squares.
Factor a sum and difference of cubes.
Factor higher power polynomials using substitution
Factor polynomials completely using all methods

Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

\begin{array}{ccc} a^{2} + 2 a b + b^{2} & = & {(a + b)}^{2} \\ and \\ a^{2} - 2 a b + b^{2} & = & {(a - b)}^{2} \end{array}

$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$

$\\$

We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

a^{2} + 2 a b + b^{2} = {(a + b)}^{2}

${a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}$

How To: Given a perfect square trinomial, factor it into the square of a binomial

Confirm that the first and last term are perfect squares.
Confirm that the middle term is twice the product of $ab$ .
Write the factored form as ${\left(a+b\right)}^{2}$ .

Example: Factoring a Perfect Square Trinomial

Factor $25{x}^{2}+20x+4$ .

Show Solution

Notice that $25{x}^{2}$ and $4$ are perfect squares because $25{x}^{2}={\left(5x\right)}^{2}$ and $4={2}^{2}$ . Then check to see if the middle term is twice the product of $5x$ and $2$ . The middle term is, indeed, twice the product: $2\left(5x\right)\left(2\right)=20x$ . Therefore, the trinomial is a perfect square trinomial and can be written as ${\left(5x+2\right)}^{2}$ .

Try It

Factor $49{x}^{2}-14x+1$ .

Show Solution

${\left(7x - 1\right)}^{2}$

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

a^{2} - b^{2} = (a + b) (a - b)

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

$\\$

We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

a^{2} - b^{2} = (a + b) (a - b)

${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

How To: Given a difference of squares, factor it into binomials

Confirm that the first and last term are perfect squares.
Write the factored form as $\left(a+b\right)\left(a-b\right)$ .

Example: Factoring a Difference of Squares

Factor $9{x}^{2}-25$ .

Show Solution

Notice that $9{x}^{2}$ and $25$ are perfect squares because $9{x}^{2}={\left(3x\right)}^{2}$ and $25={5}^{2}$ . The polynomial represents a difference of squares and can be rewritten as $\left(3x+5\right)\left(3x - 5\right)$ .

Try It

Factor $81{y}^{2}-100$ .

Show Solution

$\left(9y+10\right)\left(9y - 10\right)$

Q & A

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

Watch this video to see another example of how to factor a difference of squares.

Factoring the Sum and Difference of Cubes

Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

$\\$

Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

$\\$

We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.

x^{3} - 2^{3} = (x - 2) (x^{2} + 2 x + 4)

${x}^{3}-{2}^{3}=\left(x - 2\right)\left({x}^{2}+2x+4\right)$

The sign of the first 2 is the same as the sign between ${x}^{3}-{2}^{3}$ . The sign of the $2x$ term is opposite the sign between ${x}^{3}-{2}^{3}$ . And the sign of the last term, 4, is always positive.

A General Note: Sum and Difference of Cubes

We can factor the sum of two cubes as

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$

We can factor the difference of two cubes as

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$

How To: Given a sum of cubes or difference of cubes, factor it

Confirm that the first and last term are cubes, ${a}^{3}+{b}^{3}$ or ${a}^{3}-{b}^{3}$ .
For a sum of cubes, write the factored form as $\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$ . For a difference of cubes, write the factored form as $\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$ .

Example: Factoring a Sum of Cubes

Factor ${x}^{3}+512$ .

Show Solution

Notice that ${x}^{3}$ and $512$ are cubes because ${8}^{3}=512$ . Rewrite the sum of cubes as $\left(x+8\right)\left({x}^{2}-8x+64\right)$ .

Analysis of the Solution

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

Try It

Factor the sum of cubes $216{a}^{3}+{b}^{3}$ .

Show Solution

$\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)$

Example: Factoring a Difference of Cubes

Factor $8{x}^{3}-125$ .

Show Solution

Notice that $8{x}^{3}$ and $125$ are cubes because $8{x}^{3}={\left(2x\right)}^{3}$ and $125={5}^{3}$ . Write the difference of cubes as $\left(2x - 5\right)\left(4{x}^{2}+10x+25\right)$ .

Analysis of the Solution

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

Try It

Factor the difference of cubes: $1,000{x}^{3}-1$ .

Show Solution

$\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)$

In the following two video examples we show more binomials that can be factored as a sum or difference of cubes.

Factor Using Substitution

We are going to move back to factoring polynomials; our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor but is not quite the same. Substitution is a useful tool that can be used to “mask” a term or expression to make algebraic operations easier.

Example

Factor $x^4+3x^2+2$ .

Show Solution

This looks a lot like a trinomial that we know how to factor: $x^2+3x+2=(x+2)(x+1)$ . The only thing different is the exponents.

If we substitute $u=x^2$ and recognize that $u^2=(x^2)^2=x^4$ , we may be able to factor this beast!

Everywhere there is a $x^2$ we will replace it with a $u$ then factor.

$u^2+3u+2=(u+1)(u+2)$

We are not quite done yet. We want to factor the original polynomial which had $x$ as its variable, so we need to replace $x^2=u$ now that we are done factoring.

$(u+1)(u+2)=(x^2+1)(x^2+2)$

We conclude that $x^4+3x^2+2=(x^2+1)(x^2+2)$ .

Try It

Example

Factor $x^4-81$ .

Show Solution

This looks a lot like a difference of squares that we know how to factor: $x^2-81=(x+9)(x-9)$ . The only thing different is the exponents.

If we substitute $u=x^2$ and recognize that $u^2=(x^2)^2=x^4$ , we will be able to factor.

Everywhere there is a $x^2$ we will replace it with a $u$ then factor.

$u^2-81=(u+9)(u-9)$

We are not quite done yet. We want to factor the original polynomial which had $x$ as its variable, so we need to replace $x^2=u$ .

$(u+9)(u-9)=(x^2+9)(x^2-9)$

However, this is not fully factored. We notice that $x^2-9$ can be factored into $(x+3)(x-3)$ . You always want to factor as much as possible.

Therefore, we conclude that $x^4-81=(x^2+9)(x+3)(x-3)$ .

Try It

Factor Completely

Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example, we will first find the GCF of a trinomial, and after factoring it out, we will be able to factor again so that we end up with a product of a monomial and two binomials.

Example

Factor $6m^2k-3mk-3k$ completely.

Show Solution

Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is $3k$ .

Factor $3k$ from the trinomial:

$6m^2k-3mk-3k=3k\left(2m^2-m-1\right)$

We are left with a trinomial that can be factored using your choice of factoring methods. We will create a table to find the factors of $2\cdot{-1}=-2$ that sum to $-1$

Factors of $2\cdot-1=-2$	Sum of Factors
$2,-1$	$1$
$-2,1$	$-1$

Our factors are $-2,1$ which will allow us to factor by grouping:

Rewrite the middle term with the factors we found:

$(2m^2-m-1)=2m^2-2m+m-1$

Regroup and find the GCF of each group:

$(2m^2-2m)+(m-1)=2m(m-1)+1(m-1)$

Now factor $(m-1)$ from each term:

$2m^2-m-1=(m-1)(2m+1)$

Do not forget the original GCF that we factored out! Our final factored form is:

$6m^2k-3mk-3k=3k (m-1)(2m+1)$

Example

Factor $2x^4-144x^2+2592$ completely.

Show Solution

Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is $2$ .

Factor $2$ from the binomial:

$2(x^4-72x^2+1296)$

We are left with a trinomial that can be factored using perfect squares. Notice that $x^4=(x^2)^2$ and $(36)^2=1296$ . Also notice that the middle term, $-72x^2$ , of the trinomial is equal to $-2$ times the product of $x^2$ and $36$ . Therefore, $-2 \cdot x^2 \cdot 36=-72x^2$ . We can conclude that the trinomial $x^4-72x^2+1296$ is a perfect square trinomial.

Since the middle term is negative, use the formula $A^2-2AB+B^2=(A-B)^2$ to factor the perfect square trinomial.
$x^4-72x^2+1296=(x^2)^2-2 \cdot x^2 \cdot 36+(36)^2=(x^2-36)^2$ .

Notice that $x^2-36$ is a difference of two squares. We can factor this into $(x+6)(x-6)$ . Since this is squared, then $(x^2-36)^2=[(x+6)(x-6)]^2=(x+6)^2(x-6)^2$ .

Putting this all together, $2x^4-144x^2+2592=2(x+6)^2(x-6)^2$ .

Try It

Example

Factor $54x^{17}+2x^{14}$ completely.

Show Solution

Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is $2x^{14}$ .

Factor $2x^{14}$ from the binomial:

$2x^{14}(27x^3+1)$

We are left with a binomial that can be factored using the sum of cubes. Notice that $27x^3$ and 1 are cubes because $27x^3=(3x)^3$ and $1=1^3$ . Therefore, $27x^3+1=(3x+1)(9x^2-3x+1)$

Our final answer must include the original GCF we already factored out: $2x^{14}(3x+1)(9x^2-3x+1)$

Try It

In our last example, we show why it is important to factor out a GCF, if there is one, before you begin using the techniques shown in this section.

Chapter CR: Corequisite Review