Question

In the following exercises, factor using substitution${(x-3)}^2-5(x-3)-36$

Short answer

The factors are$\left(x-12\right)\left(x+1\right)$

Step-by-step solution

Step 1. Given information

The trinomial is ${(x-3)}^2-5(x-3)-36$.

Find factors using substitution method.

Step 2. Find the term for substitution

Rewrite the trinomial to prepare for the substitution

${(x-3)}^2-5(x-3)-36$

Substitute binomial$\left(x-3\right)=u$

So, trinomial becomes

$u^2-5u-36$

Step 3. Factor the trinomial

Split the middle term of trinomial to find factors

$$\begin{gathered} u^2-5u-36 \\ {u}^2-9u+4u-36 \end{gathered}$$

Factor it

$$\begin{gathered} u\left(u-9\right)+4\left(u-9\right) \\ \left(u-9\right)\left(u+4\right) \end{gathered}$$

Step 4. Substitute back u

Substitute back $u=x-3$

$$\begin{gathered} \left(x-3-9\right)\left(x-3+4\right) \\ \left(x-12\right)\left(x+1\right) \end{gathered}$$