Question

Factor the quadratic expression $x^2-16x+63$ completely.

Answer 1

Factor the quadratic polynomial ${\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\mathit{x}\mathbf{+}\mathbf{63}$:

Find the product of the quadratic term $x^2$ and the constant term $63$, $x^2\times 63=63x^2$ . Through a hit and trial process, we have to find two factors of $63x^2$ that add up to give the linear term $-16x$. The two factors that satisfy these conditions are $-9x$ and $-7x$, because their sum is equal to the linear term $-16x$ and their product is equal to $63x^2$.

Rewrite $-16x$ in the expression $x^2-16x+63$ as $-9x-7x$.

Then factorize by grouping the terms:

$\begin{array}{rcl}{x}^2-16x+63& =& x^2-9x-7x+63\\ & =& \left(x^2-9x\right)+\left(-7x+63\right)\\ & =& x\left(x-9\right)-7\left(x-9\right)\\ & =& \left(x-9\right)\left(x-7\right)\end{array}$

Thus, $x^2-16x+63$ is completely factorized to obtain $\mathbf{(}\mathit{x}\mathbf{-}\mathbf{9}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{7}\mathbf{)}$.