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6. Polynomials & Factoring
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Algebra 1

6.07 Factor using appropriate methods

Lesson
Practice

Factor using appropriate methods

When factoring, there are a few special products that, if we learn to recognize them, can help us factor polynomials more quickly. Recall these special products:

Perfect square trinomials

A trinomial that is made by multiplying a binomial by itself

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

Difference of two squares

Two perfect square expressions being subtracted from each other

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

Note: The sum of squares, a^{2} + b^{2}, is called prime (non-factorable).

Similar to a prime number, a prime polynomial has no factors other than 1 and itself. In other words, it has no factors with a degree less than the degree of the original polynomial.

Exploration

Consider the factored form of each polynomial expression:

Eight expressions arranged into 4 columns with 2 rows in each column. First column: left parenthesis x plus 7 right parenthesis left parenthesis x minus 7 right parenthesis, x squared minus 49; Second column: left parenthesis x minus 6 right parenthesis left parenthesis x minus 6 right parenthesis, x squared minus 12 x plus 36; Third column: left parenthesis 2 x plus 5 right parenthesis left parenthesis 2 x plus 5 right parenthesis, 4 x squared plus 20 x plus 25; Fourth column: left parenthesis x plus 4 right parenthesis left parenthesis x plus 4 right parenthesis, x squared plus 8 x plus 16.
  1. What do you notice about the factored form of the given polynomials?

When polynomials are special products, the factored form has a pattern. Knowing how the terms of the binomials relate to the terms of the expanded polynomial can help us find the factored form of these special polynomials more quickly.

When factoring polynomials, recall that the first step is to look for and factor out the greatest common factor of all terms. We can then factor by grouping, or identify a more efficient method if we recognize the patterns of special products.

For example, consider the product of a binomial squared, \left(a+b\right)^2:

An area model using algebraic tiles. Ask your teacher for more information.

\,\\\,\\\,We can expand \left(a+b\right)^2 to \left(a+b\right)\left(a+b\right) and represent them with an area model. Evaluating with this model and combining like terms, we get the product a^2 + 2ab + b^2.

So, we have:

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

Follow these steps for determining if a trinomial is a perfect square trinomial and factoring:

  1. Factor out the GCF
  2. Determine a from the leading term and b from the constant term
  3. Verify whether the linear term is equal to 2ab
  4. If yes, use the structure of perfect square trinomials to write the factors

Now consider the product of a sum and a difference, \left(a+b\right)\left(a-b\right):

Two area model using algebraic tiles. Ask your teacher for more information.

Notice that the term ab and -ab are opposites and combine to make zero. We call this a zero pair. So, (a+b)(a-b) = a^2 - b^2.

Follow these steps for determining if a binomial is a difference of two squares and factoring:

  1. Factor out the GCF
  2. Determine a from the leading term and b from the constant term
  3. Verify whether a and b are perfect squares by identifying their square roots
  4. If yes, use the structure of a difference of squares to write the factors

Examples

Example 1

Factor 3p^2+12p+12

Worked Solution
Create a strategy

We can factor a GCF of 3 out of the polynomial and write the polynomial as 3(p^2+4p+4). Determine if the polynomial expression is a perfect square trinomial.

Since a=p and b=2 and the linear term is 2ab=2(p)(2)=4p, we can verify that this is a perfect square trinomial and we can use special products to factor.

Apply the idea

Since p^2+4p+4 is a perfect square trinomial, we use the identity in factoring:

\displaystyle a^{2} + 2 a b + b^{2} \displaystyle =\displaystyle \left(a + b\right)^{2}Identity for a perfect square trinomial
\displaystyle =\displaystyle \left(p + 2\right)^{2}Substitute a=p and b=2

There are no more common factors to be divided out, so the fully factored form of the polynomial is 3(p+2)^2.

Reflect and check

If we instead were to factor by grouping, we will find the value of two integers that multiply to ac=(1)(4)=4 and add up to b=4. After finding these integers, we use them to rewrite the middle term 4p as a sum of two terms and then factor the trinomial by grouping.

The factors of 4 are 1 and 4,\,2 and 2. Among these factors, 2 and 2 are the pair that adds up to 4.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle 3(p^2+4p+4)\displaystyle =\displaystyle 3(p^2+2p+2p+4)Rewrite polynomial with four terms
\displaystyle =\displaystyle 3[p(p+2)+2(p+2)]Factor each pair
\displaystyle =\displaystyle 3(p+2)(p+2)Divide out the common factor of (p+2)

There are no more common factors to be divided out, so the fully factored form of the polynomial is 3(p+2)^2.

Example 2

Fully factor 9x^{2}−24x+16.

Worked Solution
Create a strategy

We check first whether 9x^{2}−24x+16 is a special product and identify its type.

Since a=3x and b=4 and the linear coefficient is -2ab=2(3x)(4)=-24x, we can verify that this is a perfect square trinomial and we can use special products to factor.

Apply the idea

Since 9x^{2}−24x+16 is a perfect square trinomial, we can use the identity in factoring:

\displaystyle a^{2} - 2 a b + b^{2} \displaystyle =\displaystyle \left(a - b\right)^{2}Identity for a perfect square trinomial
\displaystyle =\displaystyle \left(3x - 4\right)^{2}Substitute a=3x and b=4
Reflect and check

We can check the answer by multiplying the factored form \left(3x - 4\right)^{2}.

\displaystyle \left(3x - 4\right)^{2}\displaystyle =\displaystyle 9x^{2}−24x+16Identity for square of a binomial

Example 3

Factor 1-x^{2}.

Worked Solution
Create a strategy

We check first whether 1-x^{2} is a special product and identify its type.

Since it can be rewritten as \left(1\right)^2-\left(x\right)^2 the polynomial is a difference of squares and we can use special products to factor.

Apply the idea

Since 1-x^{2} is a difference of two squares, we use the identity in factoring:

\displaystyle a^{2} - b^{2}\displaystyle =\displaystyle \left(a+b\right)\left(a-b\right)Identity for a difference of two squares
\displaystyle =\displaystyle \left(1+x\right)\left(1-x\right)Substitute a=1 and b=x
Reflect and check

We can check the answer by multiplying the factored form \left(1+x\right)\left(1-x\right).

\displaystyle \left(1+x\right)\left(1-x\right)\displaystyle =\displaystyle 1-x^{2}Identity for product of a sum and difference

We can also check our answer using algebra tiles:

An area model with algebra tiles. Ask your teacher for more information.

So, (1-x)(1+x) is equivalent to 1-x+x-x^2 = 1-x^2.

Example 4

Factor 4m^2 + 40m +36.

Worked Solution
Create a strategy

We can factor a GCF of 4 out of the polynomial and write the polynomial as 4(m^2+10m+9). Determine if the polynomial expression is a perfect square trinomial.

Since a=m and b=3 and the linear term should be 2ab=2(m)(3)=6m and 6m \neq 10m, we can verify that this trinomial is not a perfect square trinomial and instead factor by grouping.

Apply the idea

The factors of 9 are 1 and 9,\,3 and 3. Among these factors, 1 and 9 are the factor pair that adds up to 10.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle 4(m^2+10m+9)\displaystyle =\displaystyle 4(m^2+m + 9m +9)Rewrite polynomial with four terms
\displaystyle =\displaystyle 4[m(m+1)+9(m+1)]Factor each pair
\displaystyle =\displaystyle 4(m+1)(m+9)Divide out the common factor of (m+1)

There are no more common factors to be divided out, so the fully factored form of the polynomial is 4(m+1)(m+9).

Reflect and check

If the polynomial could not be rewritten as four terms and factored by grouping, we would determine that the polynomial is not factorable.

Idea summary

By recognizing the patterns of factoring using special products, we can factor more efficiently:

  • Perfect square trinomials: a^{2} + 2 a b + b^{2} = \left(a + b\right)^{2} \text{ or } a^{2} - 2 a b + b^{2} = \left(a - b\right)^{2}
  • Difference of two squares: a^{2} - b^{2} = \left(a+b\right)\left(a-b\right)

Outcomes

A.EO.2

The student will perform operations on and factor polynomial expressions in one variable.

A.EO.2c

Factor completely first- and second-degree polynomials in one variable with integral coefficients. After factoring out the greatest common factor (GCF), leading coefficients should have no more than four factors.

A.EO.2e

Represent and demonstrate equality of quadratic expressions in different forms (e.g., concrete, verbal, symbolic, and graphical).

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