Question

In the following exercise, factor using substitution.

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

Short answer

Thus, the factorisation of the given polynomial is:

${(x-3)}^2-5(x-3)-3=(x+1)(x-12)$

Step-by-step solution

Step 1. Given information  

The given polynomial is:

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

Step 2.  substitute x-3=u 

Substituting $x-3=u$in the given polynomial, we get:

$u^2-5u-36$

Step 3. Perform factorisation 

Factorising the polynomial, we get:

$$\begin{gathered} u^2-5u-36 \\ {u}^2-9u+4u-36 \\ u(u-9)+4(u-9)[takingoutuand4ascommon] \\ (u+4)(u-9)[taking(u-9)ascommon] \end{gathered}$$

Again, substituting $u=x-3$,we get:

localid="1644654402554" $$\begin{gathered} u^2-5u-36=\left(u+4\left)\right(u-9\right) \\ {(x-3)}^2-5(x-3)-36=(x-3+4)(x-3-9) \\ {(x-3)}^2-5(x-3)-3=(x+1)(x-12) \end{gathered}$$

Thus, the factorisation of the given polynomial is:

localid="1644654442697" ${(x-3)}^2-5(x-3)-3=(x+1)(x-12)$

Step 4. Check the answer

Multiplying the factors, we get:

$$\begin{gathered} {(x-3)}^2-5(x-3)-36=(x+1)(x-12) \\ {x}^2-6x+9-5x+15-36=x^2-12x+x-12 \\ {x}^2-11x-12=x^2-11x-12 \end{gathered}$$

This is true.