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How do you factor $k^{2} + 8 k = 84$ $k^2 + 8k = 84$ ?

2 Answers

Alan P. · Meave60

Jul 29, 2015

$(k - 6) (k + 14) = 0$ $(k-6)(k+14)=0$

Explanation:

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

$k^{2} + 8 k = 84$ $k^2+8k =84$
$XXXX$ $color(white)("XXXX")$ if $k^{2} + 8 x$ $k^2+8x$ are the first 2 terms of a squared binomial)
$XXXX$ $color(white)("XXXX")$ $\to$ $rarr$ the third terms must be ${(\frac{8}{2})}^{2}$ $(8/2)^2$

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

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

$k + 4 = \pm 10$ $k+4 = +-10$

$k = 6 or k = - 14$ $k= 6 or k=-14$

$(k - 6) (k + 14) = 0$ $(k-6)(k+14) = 0$

Nghi N. · Alan P.

Jul 29, 2015

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

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

Explanation:

$y = k^{2} + 8 k - 84 .$ $y = k^2 + 8k - 84.$
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)

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