Question

Factor out the common factor.

${\left(z+2\right)}^2-5\left(z+2\right)$

Short answer

The solution${\left(z+2\right)}^2-5\left(z+2\right)=\left(z+2\right)\left(z-3\right)$

Step-by-step solution

Step 1. Apply the concept of polynomials.

Apolynomial in the variable x is an expression of the form$a_n{x}^n+a_{n-1}{x}^{n-1}+a_{n-2}{x}^{n-2}+\mathrm{...}+a_{1}x+a_0$

$a_n,a_{n-1},a_{n-2},\mathrm{...},a_1,a_0$are real numbers, and n is a nonnegative integer. If $a_n\ne 0$, then the polynomial hasdegree n. The monomials $a_k{x}^k$that make up the polynomial are called theterms of the polynomial.

Step 2. Analyze the polynomial.

Given${\left(z+2\right)}^2-5\left(z+2\right)$

Let$z+2=x$

So,${\left(z+2\right)}^2-5\left(z+2\right)=x^2-5x$

Step 3. Find the greatest common factor.

${\left(z+2\right)}^2-5\left(z+2\right)=x^2-5x$is the polynomial

1 and $-5$have greatest common factor 1

$x^2,x$have greatest common factor$x$

So, the greatest common factor of the two terms is$x$

Step 4. Factor the polynomial.

Now, the factors of polynomial${\left(z+2\right)}^2-5\left(z+2\right)=x^2-5x$are given by:

$\begin{array}{l}{\left(z+2\right)}^2-5\left(z+2\right)=x^2-5x\\ \Rightarrow {\left(z+2\right)}^2-5\left(z+2\right)=x\left(x-5\right)\\ \Rightarrow {\left(z+2\right)}^2-5\left(z+2\right)=\left(z+2\right)\left(\left(z+2\right)-5\right)\mathrm{......}\left[x=\left(z+2\right)\right]\\ \Rightarrow {\left(z+2\right)}^2-5\left(z+2\right)=\left(z+2\right)\left(z-3\right)\end{array}$