Q. 6.60 - Intermediate Algebra

Question

Factor: $x^2-6x+9-4y^2$.

Step-by-Step Solution

Answer

Factors of $x^2-6x+9-4y^2$ are $\left(x-3-2y\right)\left(x-3+2y\right)$.

Step 1. Given information.

Given an expression $x^2-6x+9-4y^2$.

Step 2. Factor by grouping the first 3 therms if it forms a perfect square trinomial.

Note that .

First 3 terms of the expression are $x^2-6x+9$ which can be expressed as ${\left(x\right)}^2-2\left(x\right)\left(3\right)+3^2$.

So, Factor by grouping as  $x^2-6x+9={\left(x-3\right)}^2$.

It follows that the expression has first 3 terms that forms a perfect square trinomial. 

Step 3. Write them as square if it is a difference of square.

It can be seen that ${\left(x-3\right)}^2-4y^2$ is a difference of square.

Consider $a=x-3$ and $b=2y$.

Step 4. Factor as products of conjugates.

Note that .

Therefore, ${\left(x-3\right)}^2-{\left(2y\right)}^2=\left(x-3-2y\right)\left(x-3+2y\right)$.