Q. 248 - Intermediate Algebra

Question

In the following exercises, factor completely.   

$4x^{5}y-32x^{2}y$

Step-by-Step Solution

Answer

The required factors are $4x^{2}y(x-2)(x^2+2x+4)$

Step 1. Given Information

The given expression is $4x^{5}y-32x^{2}y$

Step 2. Calculation

We will find the factors by factoring out the common term and then apply the property,

$$\begin{gathered} 4x^{5}y-32x^{2}y=4x^{2}y(x^3-8) \\ =4x^{2}y(x^3-2^3) \\ =4x^{2}y(x-2)(x^2+2x+4) \end{gathered}$$

Step 3. Check

We will verify the factors by multiplying the factors.

$$\begin{gathered} 4x^{2}y(x-2)(x^2+2x+4)=4x^{2}y(x^3-2^3) \\ =4x^{2}y(x^3-8) \\ =4x^{5}y-32x^{2}y \end{gathered}$$