# How do you factor k^2 + 8k = 84?

Jul 29, 2015

$\left(k - 6\right) \left(k + 14\right) = 0$

#### Explanation:

One way to approach this problem is by "completing the square"

${k}^{2} + 8 k = 84$
$\textcolor{w h i t e}{\text{XXXX}}$if ${k}^{2} + 8 x$ are the first 2 terms of a squared binomial)
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow$ the third terms must be ${\left(\frac{8}{2}\right)}^{2}$

${k}^{2} + 8 k + {4}^{2} = 84 + {4}^{2}$

${\left(k + 4\right)}^{2} = 100$

$k + 4 = \pm 10$

$k = 6 \mathmr{and} k = - 14$

$\left(k - 6\right) \left(k + 14\right) = 0$

Jul 29, 2015

Factor: y = k^2 + 8k - 84

Ans: (k - 6)(k + 14)

#### Explanation:

$y = {k}^{2} + 8 k - 84.$
I use the new AC Method. a and c have opposite signs.
Factor pairs of (-84) --> (-2, 42)(-3, 28)(-6, 14). This sum is 8 = b.

y = (k - 6)(k + 14)