Question

Factor the quadratic expression $m^2-9m+8$ completely.

Answer 1

Factor the quadratic polynomial ${\mathit{m}}^{\mathbf{2}}\mathbf{-}\mathbf{9}\mathit{m}\mathbf{+}\mathbf{8}$.

Find the product of the quadratic term $m^2$ and the constant term $8$, $m^2\times 8=8m^2$ . Through a hit and trial process, we have to find two factors of $8m^2$ that add up to give the linear term $-9m$. The two factors that satisfy these conditions are $-8m$ and $-m$, because their sum is equal to the linear term $-9m$ and their product is equal to $8m^2$.

Rewrite the term $-9m$ in the expression $m^2-9m+8$ as $-8m-m$.

Then factorize by grouping the terms:

$\begin{array}{rcl}{m}^2-9m+8& =& m^2-8m-m+8\\ & =& \left(m-8m\right)+\left(-m+8\right)\\ & =& m\left(m-8\right)-1\left(m-8\right)\\ & =& \left(m-8\right)\left(m-1\right)\end{array}$

Thus, $m^2-9m+8$ is completely factorized to obtain $\mathbf{(}\mathbf{m}\mathbf{-}\mathbf{8}\mathbf{)}\mathbf{(}\mathbf{m}\mathbf{-}\mathbf{1}\mathbf{)}$.