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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.
a2+2ab+b2=(a+b)2anda22ab+b2=(ab)2\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:
a2+2ab+b2=(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.

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

Example 4: Factoring a Perfect Square Trinomial

Factor 25x2+20x+425{x}^{2}+20x+4.

Solution

Notice that 25x225{x}^{2} and 44 are perfect squares because 25x2=(5x)225{x}^{2}={\left(5x\right)}^{2} and 4=224={2}^{2}. Then check to see if the middle term is twice the product of 5x5x and 22. The middle term is, indeed, twice the product: 2(5x)(2)=20x2\left(5x\right)\left(2\right)=20x. Therefore, the trinomial is a perfect square trinomial and can be written as (5x+2)2{\left(5x+2\right)}^{2}.

Try It 4

Factor 49x214x+149{x}^{2}-14x+1. Solution

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.
a2b2=(a+b)(ab){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.
a2b2=(a+b)(ab){a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)

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

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

Example 5: Factoring a Difference of Squares

Factor 9x2259{x}^{2}-25.

Solution

Notice that 9x29{x}^{2} and 2525 are perfect squares because 9x2=(3x)29{x}^{2}={\left(3x\right)}^{2} and 25=5225={5}^{2}. The polynomial represents a difference of squares and can be rewritten as (3x+5)(3x5)\left(3x+5\right)\left(3x - 5\right).

Try It 5

Factor 81y210081{y}^{2}-100. Solution

Q & A

Is there a formula to factor the sum of squares?

No. A sum of squares cannot be factored.

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.
a3+b3=(a+b)(a2ab+b2){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.
a3b3=(ab)(a2+ab+b2){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.
x323=(x2)(x2+2x+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 x323{x}^{3}-{2}^{3}. The sign of the 2x2x term is opposite the sign between x323{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
a3+b3=(a+b)(a2ab+b2){a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)
We can factor the difference of two cubes as
a3b3=(ab)(a2+ab+b2){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.

  1. Confirm that the first and last term are cubes, a3+b3{a}^{3}+{b}^{3} or a3b3{a}^{3}-{b}^{3}.
  2. For a sum of cubes, write the factored form as (a+b)(a2ab+b2)\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right). For a difference of cubes, write the factored form as (ab)(a2+ab+b2)\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right).

Example 6: Factoring a Sum of Cubes

Factor x3+512{x}^{3}+512.

Solution

Notice that x3{x}^{3} and 512512 are cubes because 83=512{8}^{3}=512. Rewrite the sum of cubes as (x+8)(x28x+64)\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 6

Factor the sum of cubes: 216a3+b3216{a}^{3}+{b}^{3}. Solution

Example 7: Factoring a Difference of Cubes

Factor 8x31258{x}^{3}-125.

Solution

Notice that 8x38{x}^{3} and 125125 are cubes because 8x3=(2x)38{x}^{3}={\left(2x\right)}^{3} and 125=53125={5}^{3}. Write the difference of cubes as (2x5)(4x2+10x+25)\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 7

Factor the difference of cubes: 1,000x311,000{x}^{3}-1. Solution

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