When factoring there are a few special products that, if we learn to recognize, can help us factor polynomials more quickly. Two of these special products are perfect square trinomials and difference of two squares.

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 always prime (non-factorable).

Note: To factor some polynomials completely, we may need to combine different factoring methods (finding common factors, grouping, and using special products).

Worked examples

Example 1

Factor x^{2} + 6 x + 9.

Approach

We check first whether x^{2} + 6 x + 9 is a special product and identify its type.

Solution

Since x^{2} + 6 x + 9 is a perfect square trinomial, we use the formula in factoring:

\displaystyle a^{2} + 2 a b + b^{2}	\displaystyle =	\displaystyle \left(a + b\right)^{2}	Formula of a perfect square trinomial
\displaystyle x^{2} + 6x + 9	\displaystyle =	\displaystyle x^{2} + 2(3x) + 3^{2}	Substitute the given values
	\displaystyle =	\displaystyle \left(x + 3\right)^{2}	Final answer

Reflection

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

\displaystyle \left(x + 3\right)^{2}

\displaystyle =

\displaystyle x^{2} + 6 x + 9

Square of a binomial

Example 2

Fully factor 9x^{2}−24x+16.

Approach

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

Solution

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

\displaystyle a^{2} - 2 a b + b^{2}	\displaystyle =	\displaystyle \left(a - b\right)^{2}	Formula of a perfect square trinomial
\displaystyle 9x^{2}−24x+16	\displaystyle =	\displaystyle \left(3x\right)^{2}−2(4)(3x)+4^{2}	Substitute the given values
	\displaystyle =	\displaystyle \left(3x - 4\right)^{2}	Final answer

Reflection

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+16

Square of a binomial

Example 3

Factor x^{2} - 25.

Approach

We check first whether x^{2} - 25 is a special product and identify its type.

Solution

Since x^{2} - 25 is a difference of two squares, we use the formula in factoring:

\displaystyle a^{2} - b^{2}	\displaystyle =	\displaystyle \left(a+b\right)\left(a-b\right)	Formula of difference of two squares
\displaystyle x^{2} - 25	\displaystyle =	\displaystyle x^{2} - 5^{2}	Substitute the given values
	\displaystyle =	\displaystyle \left(x+5\right)\left(x-5\right)	Final answer

Reflection

We can check the answer by multiplying the factored form \left(x+5\right)\left(x-5\right).

\displaystyle \left(x+5\right)\left(x-5\right)

\displaystyle =

\displaystyle x^{2} - 25

Product of a sum and difference

Example 4

Factor 3 x^{2} - 27.

Approach

We find a common factor first then check whether the polynomial is a special product.

Solution

The common factor for the two terms is 3. We factor out 3 and get 3\left(x^{2} - 9\right).

Since x^{2} - 9 is a difference of two squares, we use the formula in factoring:

\displaystyle a^{2} - b^{2}	\displaystyle =	\displaystyle \left(a+b\right)\left(a-b\right)	Formula of difference of two squares
\displaystyle x^{2} - 9	\displaystyle =	\displaystyle x^{2} - 3^{2}	Substitute the given values
	\displaystyle =	\displaystyle \left(x+3\right)\left(x-3\right)	Factors of x^{2} - 9

Adding the common factor 3, the final answer is 3\left(x+3\right)\left(x-3\right).

Reflection

We can check the answer by multiplying the factored form 3\left(x+3\right)\left(x-3\right).

\displaystyle 3\left(x+3\right)\left(x-3\right)	\displaystyle =	\displaystyle 3\left(x^{2} - 9\right)	Product of a sum and difference
	\displaystyle =	\displaystyle 3x^{2} - 27	Distribute 3

Outcomes

A1.A.APR.A.1

Add, subtract, and multiply polynomials. Use these operations to demonstrate that polynomials form a closed system that adhere to the same properties of operations as the integers.

6.09 Factoring special products

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Example 4

Approach

Solution

Reflection

Outcomes

A1.A.APR.A.1

A1.MP1

A1.MP4

A1.MP7

A1.MP8

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