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

1 Answer
George C.
Aug 19, 2017

#14x^2-7x-28 = 7/8(4x-1-sqrt(33))(4x-1+sqrt(33))#

Explanation:

Given:

#f(x) = 14x^2-7x-28#

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

#8/7 f(x) = 16x^2-8x-32#

#color(white)(8/7 f(x)) = (4x)^2-2(4x)+1-33#

#color(white)(8/7 f(x)) = (4x-1)^2-(sqrt(33))^2#

#color(white)(8/7 f(x)) = ((4x-1)-sqrt(33))((4x-1)+sqrt(33))#

#color(white)(8/7 f(x)) = (4x-1-sqrt(33))(4x-1+sqrt(33))#

Multiplying both ends by #7/8# we find:

#14x^2-7x-28 = 7/8(4x-1-sqrt(33))(4x-1+sqrt(33))#

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