How do you factor #14x^2+8x+72#?

1 Answer
Apr 13, 2017

Answer:

This quadratic has no factors with real coefficients, but we find:

#14x^2+8x+72 = 2/7(7x+2-2sqrt(62)i)(7x+2+2sqrt(62)i)#

Explanation:

This quadratic is not factorisable using real coefficients.

We can factor it using complex coefficients by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)

with #a=(7x+2)# and #b=2sqrt(62)# as follows:

#(7/2)(14x^2+8x+72) = 49x^2+28x+252#

#color(white)((7/2)(14x^2+8x+72)) = (7x)^2+2(7x)(2)+2^2+248#

#color(white)((7/2)(14x^2+8x+72)) = (7x+2)^2+(2sqrt(62))^2#

#color(white)((7/2)(14x^2+8x+72)) = (7x+2)^2-(2sqrt(62)i)^2#

#color(white)((7/2)(14x^2+8x+72)) = ((7x+2)-2sqrt(62)i)((7x+2)+2sqrt(62)i)#

#color(white)((7/2)(14x^2+8x+72)) = (7x+2-2sqrt(62)i)(7x+2+2sqrt(62)i)#

So:

#14x^2+8x+72 = 2/7(7x+2-2sqrt(62)i)(7x+2+2sqrt(62)i)#