Q. 252 - Intermediate Algebra

Question

In the following exercises, factor completely.   

$48x^5{y}^2-243x{y}^2$

Step-by-Step Solution

Answer

The required factors are $3x{y}^2(4x^2+9)(2x+3)(2x-3)$

Step 1. Given Information

The given expression is $48x^5{y}^2-243x{y}^2$

Step 2. Calculation

We will find the factors by using difference of square patterns,

$$\begin{gathered} 48x^5{y}^2-243x{y}^2=3x{y}^2(16x^4-81) \\ =3x{y}^2({\left(4x^2\right)}^2-9^2) \\ =3x{y}^2(4x^2+9)(4x^2-9) \\ =3x{y}^2(4x^2+9)({\left(2x\right)}^2-3^2) \\ =3x{y}^2(4x^2+9)(2x+3)(2x-3) \end{gathered}$$

Step 3. Check

We will verify the factors by multiplying the factors.

$$\begin{gathered} 3x{y}^2(4x^2+9)(2x+3)(2x-3)=3x{y}^2(4x^2+9)({\left(2x\right)}^2-3^2) \\ =3x{y}^2(4x^2+9)(4x^2-9) \\ =3x{y}^2({\left(4x^2\right)}^2-9^2) \\ =3x{y}^2(16x^4-81) \\ =48x^5{y}^2-243x{y}^2 \end{gathered}$$