1.6 Factoring polynomials: factoring two special products (Page 2/2)

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$3 x^{2} - 75$

$3 (x 5) (x 5)$

$a^{3} b^{4} m a m^{3} n^{2}$

$a m (a b^{2} + m n) (a b^{2} - m n)$

Fundamental rules of factoring

There are two fundamental rules that we follow when factoring:

Fundamental rules of factoring

Factor out all common monomials first.
Factor completely.

Sample set b

Factor each binomial completely.

$\begin{matrix} 4 a^{8} b - 36 b^{5} . & Factor out the common factor 4b. \\ 4 b (a^{8} - 9 b^{4}) \end{matrix}$

Now we can see a difference of two squares, whereas in the original polynomial we could not. We’ll complete our factorization by factoring the difference of two squares.

$\begin{array}{l} 4 a^{8} b - 36 b^{5} & = & 4 b (a^{8} - 9 b^{4}) \\ = & 4 b (a^{4} + 3 b^{2}) (a^{4} - 3 b^{2}) \end{array}$

$\begin{array}{l} x^{16} - y^{8} . & Factor this difference of two squares . \\ x^{16} - y^{8} & = & \underset{\begin{matrix} Sum of two squares \\ Does not factor \end{matrix}}{\underset{}{(x^{8} + y^{4})}} \underset{\begin{matrix} Difference of two squares \\ Factor it! \end{matrix}}{\underset{}{(x^{8} - y^{4})}} \\ = & (x^{8} + y^{4}) (x^{4} + y^{2}) \underset{Factor again!}{\underset{}{(x^{4} - y^{2})}} \\ = & (x^{8} + y^{4}) (x^{4} + y^{2}) (x^{2} + y) (x^{2} - y) \end{array}$

Finally, the factorization is complete.

These types of products appear from time to time, so be aware that you may have to factor more than once.

Practice set b

Factor each binomial completely.

$m^{4} - n^{4}$

$(m^{2} + n^{2}) (m - n) (m + n)$

$16 y^{8} - 1$

$(4 y^{4} + 1) (2 y^{2} + 1) (2 y^{2} - 1)$

Perfect square trinomials

Recall the process of squaring a binomial.

${(a b)}^{2} a^{2} 2 a b b^{2}$ ${(a b)}^{2} a^{2} 2 a b b^{2}$

Our Method Is	We Notice
Square the first term.	The first term of the product should be a perfect square.
Take the product of the two terms and double it.	The middle term of the product should be divisible by 2 (since it’s multiplied by 2).
Square the last term.	The last term of the product should be a perfect square.

Perfect square trinomials always factor as the square of a binomial.

To recognize a perfect square trinomial, look for the following features:

The first and last terms are perfect squares.
The middle term is divisible by 2, and if we divide the middle term in half (the opposite of doubling it), we will get the product of the terms that when squared produce the first and last terms.

In other words, factoring a perfect square trinomial amounts to finding the terms that, when squared, produce the first and last terms of the trinomial, and substituting into one of the formula

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

Sample set c

Factor each perfect square trinomial.

$x^{2} + 6 x + 9$ . This expression is a perfect square trinomial. The $x^{2}$ and 9 are perfect squares.

The terms that when squared produce $x^{2}$ and 9 are $x$ and 3, respectively.

The middle term is divisible by 2, and $\frac{6 x}{2} 3 x$ . The $3 x$ is the product of $x$ and 3, which are the terms that produce the perfect squares.

$x^{2} + 6 x + 9 = {(x + 3)}^{2}$

$x^{4} - 10 x^{2} y^{3} + 25 y^{6}$ . This expression is a perfect square trinomial. The $x^{4}$ and $25 y^{6}$ are both perfect squares. The terms that when squared produce $x^{4}$ and $25 y^{6}$ are $x^{2}$ and $5 y^{3}$ , respectively.

The middle term $- 10 x^{2} y^{3}$ is divisible by $2$ . In fact, $\frac{- 10 x^{2} y^{3}}{2} = - 5 x^{2} y^{3}$ . Thus,

$x^{4} - 10 x^{2} y^{3} + 25 y^{6} = {(x^{2} - 5 y^{3})}^{2}$

$x^{2} + 10 x + 16$ . This expression is not a perfect square trinomial. Although the middle term is divisible by $2, ? \frac{10 x}{2} = 5 x$ , the 5 and $x$ are not the terms that when squared produce the first and last terms. (This expression would be a perfect square trinomial if the middle term were $8 x$ .)

$4 a^{4} + 32 a^{2} b - 64 b^{2}$ . This expression is not a perfect square trinomial since the last term $- 64 b^{2}$ is not a perfect square (since any quantity squared is always positive or zero and never negative).

Thus, $4 a^{4} + 32 a^{2} b - 64 b^{2}$ cannot be factored using this method.

Practice set c

Factor, if possible, the following trinomials.

$m^{2} - 8 m + 16$

${(m - 4)}^{2}$

$k^{2} + 10 k + 25$

${(k + 5)}^{2}$

$4 a^{2} + 12 a + 9$

${(2 a + 3)}^{2}$

$9 x^{2} - 24 x y + 16 y^{2}$

${(3 x - 4 y)}^{2}$

$2 w^{3} z + 16 w^{2} z^{2} + 32 w z^{3}$

$2 w z {(w + 4 z)}^{2}$

$x^{2} + 12 x + 49$

not possible

Exercises

For the following problems, factor the binomials.

$a 2 9$

$(a + 3) (a - 3)$

$a 2 25$

$x 2 16$

$(x + 4) (x - 4)$

$y 2 49$

$a 2 100$

$(a + 10) (a - 10)$

$b 2 36$

$4 a 2 64$

$4 (a + 4) (a - 4)$

$2 b 2 32$

$3 x 2 27$

$3 (x + 3) (x - 3)$

$5 x 2 125$

$4 a 2 25$

$(2 a + 5) (2 a - 5)$

$9 x 2 100$

$36 y 2 25$

$(6 y + 5) (6 y - 5)$

$121 a 2 9$

$12 a 2 75$

$3 (2 a + 5) (2 a - 5)$

$10 y 2 320$

$8 y 2 50$

$2 (2 y + 5) (2 y - 5)$

$a 2 b 2 9$

$x 2 y 2 25$

$(x y + 5) (x y - 5)$

$x 4 y 4 36$

$x 4 y 4 9 a 2$

$(x^{2} y^{2} + 3 a) (x^{2} y^{2} - 3 a)$

$a 2 b 4 16 y 4$

$4 a 2 b 2 9 b 2$

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

$16 x^{2} 25 y^{2}$

$a^{2} b^{2}$

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

$a^{4} b^{4}$

$x^{4} y^{4}$

$(x^{2} + y^{2}) (x + y) (x - y)$

$x^{8} y^{2}$

$a^{8} y^{2}$

$(a^{4} + y) (a^{4} - y)$

$b^{6} y^{2}$

$b^{6} x^{4}$

$(b^{3} + x^{2}) (b^{3} - x^{2})$

$9 x^{2}$

$25 a^{2}$

$(5 + a) (5 - a)$

$49 16 a^{2}$

$100 36 b^{4}$

$4 (5 + 3 b^{2}) (5 - 3 b^{2})$

$128 32 x^{2}$

$x 4 16$

$(x^{2} + 4) (x + 2) (x - 2)$

$2 a b^{3} - a^{3} b$

$a^{4} b^{4}$

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

$a^{16} b^{4}$

$x^{12} y^{12}$

$(x^{6} + x^{6}) (x^{3} + y^{3}) (x^{3} - y^{3})$

$a^{2} c 9 c$

$a^{3} c^{2} - 25 a c^{2}$

$a c^{2} (a + 5) (a - 5)$

$a^{4} b^{4} c^{2} d^{2} - 36 x^{2} y^{2}$

$49 x^{2} y^{4} z^{6} - 64 a^{4} b^{2} c^{8} d^{10}$

$(7 x y^{2} z^{3} + 8 a^{2} b c^{4} d^{5}) (7 x y^{2} z^{3} - 8 a^{2} b c^{4} d^{5})$

For the following problems, factor, if possible, the trinomials.

$x^{2} 8 x 16$

$x^{2} 10 x 25$

${(x + 5)}^{2}$

$a^{2} 4 a 4$

$a^{2} 12 a 36$

${(a + 6)}^{2}$

$b^{2} 18 b 81$

$y^{2} 20 y 100$

${(y + 10)}^{2}$

$c^{2} 6 c 9$

$a^{2} 4 a 4$

${(a - 2)}^{2}$

$b^{2} 6 b 9$

$x^{2} 10 x 25$

${(x - 5)}^{2}$

$b^{2} 22 b 121$

$a^{2} 24 a 144$

${(a - 12)}^{2}$

$a^{2} 2 a 1$

$x^{2} 2 x 1$

${(x + 1)}^{2}$

$x^{2} 2 x 1$

$b^{2} 2 b 1$

${(b - 1)}^{2}$

$4 a^{2} 12 a 9$

$9 x^{2} 6 x 1$

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

$4 x^{2} 28 x 49$

$16 a^{2} 24 a 9$

${(4 a - 3)}^{2}$

$25 a^{2} 20 a 4$

$9 x^{2} 6 x y y^{2}$

${(3 x + y)}^{2}$

$16 x^{2} 24 x y 9 y^{2}$

$36 a^{2} 60 a b 25 b^{2}$

${(6 a + 5 b)}^{2}$

$4 x^{2} 12 x y 9 y^{2}$

$12 a^{2} 60 a 75$

$3 {(2 a - 5)}^{2}$

$16 x^{2} 8 x 1$

$32 x^{2} 16 x 2$

$2 {(4 x + 1)}^{2}$

$x^{2} x 1$

$4 a^{2} a 9$

not factorable

$9 a^{2} 21 a 49$

$x^{5} 8 x^{4} 16 x^{3}$

$x^{3} {(x + 4)}^{2}$

$12 a^{3} b - 48 a^{2} b^{2} + 48 a b^{3}$

Exercises for review

( [link] ) Factor $(m 3) x (m 3) y$ .

$(m - 3) (x - y)$

( [link] ) Factor $8 x m 16 x n 3 y m 6 y n$ by grouping.

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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