Chapter 8.5, Problem 19PPE

To determine

To calculate: The factor of $q^2-8q+12$

Answer to Problem 19PPE

  $(q-2)(q-6)$

Explanation of Solution

Given information: The expression $q^2-8q+12$

Calculation:

  $\begin{array}{l}{q}^2-8q+12\\ =q^2-(6+2)q+12[\because 12=6\times 2\text{and}\text{ 8}=6+2]\\ =q^2-6q-2q+12\\ =(q^2-6q)-(2q-12)\\ =q(q-6)-2(q-6)\\ =(q-2)(q-6)\end{array}$

To check,

  $\begin{array}{l}(q-2)(q-6)\\ =q^2-6q-2q+12[\text{by FOIL method]}\\ =q^2-8q+12\end{array}$

Thus, $q^2-8q+12=(q-2)(q-6)$