Q. 421 - Intermediate Algebra

Question

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

$16x^2-24xy+9y^2-64$

Step-by-Step Solution

Answer

The factors of the given equation are $(4x-3y+8)(4x-3y-8)$.

Step 1. Given and explanation.

We have $16x^2-24xy+9y^2-64$.

We will first group the first three terms as it forms the identity of ${(a-b)}^2=a^2-ab+b^2$ .

Then further we will find the property of ${\left(a\right)}^2-{\left(b\right)}^2=(a+b)(a-b)$. So by applying this we will get the factors for the equation.

Step 2. Simplifying the equation.

We have $(16x^2-24xy+9y^2)-64$.

Here, if we assume $a=4x$and 
$b=3y$, this forms the identity of ${(a-b)}^2$. This can be written as ${(4x-3y)}^2-64$.

We know that ${\left(8\right)}^2=64$, so this forms the property of ${\left(a\right)}^2-{\left(b\right)}^2=(a+b)(a-b)$.

This gives us,

$={(4x-3y)}^2-64={(4x-3y)}^2-(8{)}^2=(4x-3y+8\left)\right(4x-3y+8)$

This cannot be further factored so these are the factors of the solution.

Step 3. 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.

We have $(4x-3y+8)(4x-3y-8)$.

So,

$$\begin{gathered} =(4x-3y+8)(4x-3y-8) \\ =16x^2-12xy-32x-12xy+9y^2+24y+32x-24y-64 \\ =16x^2-24xy+9y^2-64 \end{gathered}$$

Thus our calculations are right.