How do you factor $14 x^{2} - 7 x - 28$ $14x ^ { 2} - 7x - 28$ ?

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How do you factor $14 x^{2} - 7 x - 28$ $14x ^ { 2} - 7x - 28$ ?

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Answer 1 · 2017-08-19T20:45:34

$14 x^{2} - 7 x - 28 = \frac{7}{8} (4 x - 1 - \sqrt{33}) (4 x - 1 + \sqrt{33})$ $14x^2-7x-28 = 7/8(4x-1-sqrt(33))(4x-1+sqrt(33))$

Explanation:

Given:

$f (x) = 14 x^{2} - 7 x - 28$ $f(x) = 14x^2-7x-28$

I will multiply by $\frac{8}{7}$ $8/7$ in order to complete the square and use the difference of squares identity while avoiding fractions as much as possible...

$\frac{8}{7} f (x) = 16 x^{2} - 8 x - 32$ $8/7 f(x) = 16x^2-8x-32$

$\frac{8}{7} f (x) = {(4 x)}^{2} - 2 (4 x) + 1 - 33$ $color(white)(8/7 f(x)) = (4x)^2-2(4x)+1-33$

$\frac{8}{7} f (x) = {(4 x - 1)}^{2} - {(\sqrt{33})}^{2}$ $color(white)(8/7 f(x)) = (4x-1)^2-(sqrt(33))^2$

$\frac{8}{7} f (x) = ((4 x - 1) - \sqrt{33}) ((4 x - 1) + \sqrt{33})$ $color(white)(8/7 f(x)) = ((4x-1)-sqrt(33))((4x-1)+sqrt(33))$

$\frac{8}{7} f (x) = (4 x - 1 - \sqrt{33}) (4 x - 1 + \sqrt{33})$ $color(white)(8/7 f(x)) = (4x-1-sqrt(33))(4x-1+sqrt(33))$

Multiplying both ends by $\frac{7}{8}$ $7/8$ we find:

$14 x^{2} - 7 x - 28 = \frac{7}{8} (4 x - 1 - \sqrt{33}) (4 x - 1 + \sqrt{33})$ $14x^2-7x-28 = 7/8(4x-1-sqrt(33))(4x-1+sqrt(33))$

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How do you factor $14 x^{2} - 7 x - 28$ $14x ^ { 2} - 7x - 28$ ?

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