Factor by substitution:

${(x-5)}^2+6(x-5)+8$

We have an equation ${(x-5)}^2+6(x-5)+8$.

We can see that the binomial in the middle term is squared in the first term. So, if we substitute this binomial with a variable, we can get an equation in the standard form and it will then be easy for us to factorize the equation.

The replacing the variable with the original binomials nd simplifying it will give us the factors.

We have ${(x-5)}^2+6(x-5)+8$.

Replacing the binomial by variable $"u"$gives us $u^2+6u+8$.

We have the new equation as $u^2+6u+8$.

We need two numbers whose sum comes out to be the product of first and last term, i.e., $8$and the sum comes out to be the middle term, i.e., $6$. The factors of $8$are-

This shows that the two numbers are$2,4$.

We have two numbers as $2,4$.

Splitting the middle term gives us,

$$\begin{gathered} =u^2+6u+8 \\ =u^2+2u+4u+8 \\ =u(u+2)+4(u+2) \\ =(u+2)(u+4) \end{gathered}$$

Replacing $u$by $(x-5)$gives us

$$\begin{gathered} =(u+2)(u+4) \\ =\left[(x-5)+2\right]\left[(x-5)+4\right] \\ =(x-3)(x-1) \end{gathered}$$

This shows that factors are$(x-3)(x-1)$.