How do you factor #6x^2+23x+6#?

2 Answers
Binayaka C.
Jun 4, 2016

Root1 #= - 0.282 (3dp)# & Root2 #= -3.552 (3dp)#

Explanation:

This is a quadratic equation of form #ax^2+bx+c# Here #a=6;b=23;c=6# Roots are #-b/(2a)+- sqrt(b^2-4ac)/(2a) or -23/12+-sqrt(529-144)/12 = -1.916 +- 19.62/12 = -1.916 +- 1.635 :.# Root1 #= - 0.282# & Root2 #= -3.552#[Ans]

George C.
Jun 4, 2016

#6x^2+23x+6=6(x+(23-sqrt(385))/12)(x+(23+sqrt(385))/12)#

Explanation:

Complete the square, then use the difference of squares identity:

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

with #a=(x+23/12)# and #b=sqrt(385)/12# as follows:

#6x^2+23x+6#

#=6(x^2+23/6x+1)#

#=6((x+23/12)^2-23^2/12^2+1)#

#=6((x+23/12)^2-(529-144)/12^2)#

#=6((x+23/12)^2-385/12^2)#

#=6((x+23/12)-sqrt(385)/12)((x+23/12)+sqrt(385)/12)#

#=6(x+(23-sqrt(385))/12)(x+(23+sqrt(385))/12)#

Note that it is not possible to simplify #sqrt(385)# any further since #385 = 5*7*11# has no square factors.

Related questions
See all questions in Factoring Completely
Impact of this question
1689 views around the world
You can reuse this answer
Creative Commons License