Q 270 - Intermediate Algebra

Question

In the following exercises, factor completely.  

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

Step-by-Step Solution

Answer

The expression is factored as $16x^2-24xy+9y^2-64=(4x-3y-8)(4x-3y+8)$

Step 1. Given Information

Consider the expression $16x^2-24xy+9y^2-64$.

The objective is to factor the expression completely.

Step 2. Factor using the perfect square trinomial

The trinomial $a^2-2ab+b^2$ is a perfect square trinomial and it is factored as ${(a-b)}^2$.

In the given expression, the first three terms form a perfect square trinomial and so it can be factored as 

$\begin{array}{rcl}16x^2-24xy+9y^2-64& =& {\left(4x\right)}^2-2·4x·3y+{\left(3y\right)}^2-64 \\ & =& {(4x-3y)}^2-64\end{array}$

Step 3. Factor using the difference of square

The difference of squares can be factored as $a^2-b^2=(a-b)(a+b)$

Now the first term is a perfect square and the second term $64$ is the square of $8$. So it can be further factored as

$\begin{array}{rcl}{(4x-3y)}^2-64& =& {(4x-3y)}^2-8^2 \\ & =& (4x-3y-8)(4x-3y+8)\end{array}$

Step 4. Check the factors

Multiply the factors to check the solution 

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

And we get the given expression. So the expression is correctly factored.