How do you factor #6x^2+7x-62#?

1 Answer
Nov 19, 2016

Answer:

#f(x) = 6(x+7/12-sqrt(1537)/12)(x+7/12+sqrt(1537)/12)#

Explanation:

The difference of squares identity can be written:

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

We use this with #a=(12x+7)# and #b=sqrt(1537)# later.

Given:

#f(x) = 6x^2+7x-62#

#24f(x) = 144x^2+168x-1488#

#color(white)(24f(x)) = (12x+7)^2-49-1488#

#color(white)(24f(x)) = (12x+7)^2-1537#

#color(white)(24f(x)) = (12x+7)^2-(sqrt(1537))^2#

#color(white)(24f(x)) = ((12x+7)-sqrt(1537))((12x+7)+sqrt(1537))#

#color(white)(24f(x)) = (12x+7-sqrt(1537))(12x+7+sqrt(1537))#

#color(white)(24f(x)) = 144(x+7/12-sqrt(1537)/12)(x+7/12+sqrt(1537)/12)#

Hence:

#f(x) = 6(x+7/12-sqrt(1537)/12)(x+7/12+sqrt(1537)/12)#