1.5 Factoring polynomials (Page 3/6)

College algebra

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Factor a. $2 x^{2} + 9 x + 9$ b. $6 x^{2} + x - 1$

a. $(2 x + 3) (x + 3)$ b. $(3 x −1) (2 x + 1)$

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{matrix} a^{2} + 2 a b + b^{2} & = & {(a + b)}^{2} \\ and \\ a^{2} - 2 a b + b^{2} & = & {(a - b)}^{2} \end{matrix}

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}

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 $a b .$
Write the factored form as ${(a + b)}^{2} .$

Factoring a perfect square trinomial

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

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

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

${(7 x −1)}^{2}$

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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)

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)

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 $(a + b) (a - b) .$

Factoring a difference of squares

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

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

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

$(9 y + 10) (9 y - 10)$

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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.

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

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})

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: S ame O pposite A lways P ositive. For example, consider the following example.

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

The sign of the first 2 is the same as the sign between $x^{3} - 2^{3} .$ The sign of the $2 x$ 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})

We can factor the difference of two cubes as

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

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Source: OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3

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