91Ó°ÊÓ

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.

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.

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,

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