Step 1
Given;
[tex](x-3)(x^2+4x-11)+(x-3)^2_{}[/tex]Required; To factorize the problem
Step 2
Find the GCF
[tex]\text{The GCF is x-3 because it is common}[/tex]Hence, we will factorize thus and have;
[tex]\text{GCF(}\frac{Total\text{ first term}}{\text{GCF}}+\frac{Total\text{ second term}}{\text{GCF}})[/tex]Applying this we will have;
[tex]x-3(\frac{(x-3)(x^2+4x-11)}{x-3}+\frac{(x-3)^2}{x-3})[/tex][tex]\begin{gathered} (x-3)((x^2+4x-11)_{}+(x-3)) \\ (x-3)(x^2+4x+x-11-3)--\text{ open bracket and add like terms} \\ (x-3)(x^2+5x-14) \end{gathered}[/tex]Therefore, we will now factorize (x²+5x-14) by finding the terms that when added together gives 5x and when multiplied together gives -14x²
[tex]\begin{gathered} \text{These factors are 7x and -2x} \\ \end{gathered}[/tex]Replace 5x with (7x-2x)
[tex]x^2+7x-2x-14[/tex]Grouping the four terms in twos, that is the first two as a pair and the remaining two as the other pair, we can bring out common terms from each pair. This is shown below:
[tex]\begin{gathered} x^2\text{ }and\text{ }7x\Rightarrow x \\ -2x\text{ }and-14\Rightarrow-2 \end{gathered}[/tex]Therefore, using the factoring method shown in Step 2, we have the expression to be:
[tex]\begin{gathered} x^2+7x-2x-14 \\ \text{Put the brackets} \\ (x^2+7x)(-2x-14) \\ \text{find the GCF in each bracket} \\ In\text{ the first bracket the GCF is x. In the second bracket the GCF is -}2 \\ x(\frac{x^2}{x}+\frac{7x}{x})-2(\frac{-2x}{-2}-\frac{14}{-2}) \\ \text{divide by GCF in each bracket} \\ x(x+7)-2(x+7) \end{gathered}[/tex]Collecting the like terms, we, therefore, have the factorized expression to be:
[tex]\Rightarrow(x+7)(x-2)[/tex]Therefore the full factorized expression will be;
[tex](x-3)(x+7)(x-2)[/tex]The answer will therefore be ; (x-3)(x+7)(x-2)
