Q. 416 - Intermediate Algebra

Question

Recognize and Use the Appropriate Method to Factor a Polynomial Completely.

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

Step-by-Step Solution

Answer

The factors of the given expression are $3x{y}^2(4x^2+9)(2x+3)(2x-3)$.

Step 1. Given and explanation.

We have $48x^5{y}^2-243x{y}^2$.

We will have to first take the common factor out from the equation. We will then find a pattern of difference of two perfect squares, i.e., ${\left(a\right)}^2-{\left(b\right)}^2=(a+b)(a-b)$.

This will give us the factors.

Step 2. Taking out the common factor and simplifying.

We have $48x^5{y}^2-243x{y}^2$.

We can take $3x{y}^2$ common and it will give us,

$=3x{y}^2(16x^4-81)$.

We can see that ${\left(4x^2\right)}^2=16x^4$ and
${\left(9\right)}^2=81$.

This is the property of difference of two perfect cubes.

Step 3. Factorizing and simplifying.

We have $3x{y}^2(16x^4-81)$.

We know that they are perfect cubes, so applying that property will give us,

$=3x{y}^2(4x^2+9)(4x^2-9)$

We can still apply this property in $(4x^2-9)$ as ${\left(2x\right)}^2=4x^2$ and
${\left(3\right)}^2=9$.

So further simplifying it gives us,

$=3x{y}^2(4x^2+9)(2x+3)(2x-3)$.

This cannot be further factored. So, these are the factors of the given expression.

Step 4. Check the solution

We will check the solution by simply multiplying the factors. If we get the given equation as the product, our answer is right.

So,

$$\begin{gathered} =3x{y}^2(4x^2+9)(2x+3)(2x-3) \\ =\left[12x^3{y}^2+27x{y}^2\right]\left[4x^2-9\right] \\ =48x^5{y}^2-108x^3{y}^2+108x^3{y}^2-243x{y}^2 \\ =48x^5{y}^2-243x{y}^2 \end{gathered}$$

Thus our calculations are right.