Q 6.59 - Intermediate Algebra

Question

Factor: $x^2-10x+25-y^2$.

Step-by-Step Solution

Answer

Factors of $x^2-10x+25-y^2$ are $\left(x-5-y\right)\left(x-5+y\right)$.

Step 1. Given information.

Given an expression $x^2-10x+25-y^2$.

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

Note that $a^2-2ab+b^2={\left(a-b\right)}^2$.

First 3 terms of the expression are $x^2-10x+25$ which can be expressed as ${\left(x\right)}^2-2\left(x\right)\left(5\right)+5^2$.

So, Factor by grouping as  $x^2-10x+25={\left(x-5\right)}^2$.

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 ${\left(x-5\right)}^2-y^2$ is a difference of square.

Consider $a=x-5$ and $b=y$.

Step 4. Factor as products of conjugates.

Note that $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Therefore, ${\left(x-5\right)}^2-y^2=\left(x-5-y\right)\left(x-5+y\right)$.