Q 6.57 - Intermediate Algebra

Question

Factor: $2x^4{y}^2-32y^2$.

Step-by-Step Solution

Answer

Factors of $2x^4{y}^2-32y^2$ are $2y^2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)$.

Step 1. Given information.

Given an expression $2x^4{y}^2-32y^2$.

Step 2. Find the GCF of the expression.

Note that GCF stands for Greatest common factor.

In the expression $2x^4{y}^2-32y^2$ has $2y^2$ as its GCF.

The expression can be written as $2y^2\left(x^4-16\right)$.

Step 3. Find if the binomial is a difference of squares and factor as a product of conjugates.

Note that $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

The binomial can be expressed as a difference of squares as follows.

$\begin{array}{rcl}2y^2\left(x^4-16\right)& =& 2y^2\left({\left(x^2\right)}^2-4^2\right)\\ & =& 2y^2\left(x^2-4\right)\left(x^2+4\right)\\ & & \end{array}$

Step 4. Check if the factors are also a difference of squares.

It can be seen that factor $x^2-4$ is a difference of squares.

Note that sum of squares cannot be factored.

Factor the result as the product of conjugates as follows.

$\begin{array}{rcl}2y^2\left(x^2-4\right)\left(x^2+4\right)& =& 2y^2\left(x^2-2^2\right)\left(x^2+4\right)\\ & =& 2y^2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\end{array}$

Step 5. Verify.

Multiply the factors as follows.

$\begin{array}{rcl}2y^2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)& =& 2y^2\left(x^2-2^2\right)\left(x^2+4\right)\\ & =& 2y^2\left(x^4-2^4\right)\\ & =& 2x^4{y}^2-32y^2\end{array}$