How do you factor #4x^2-8x-12+6x#?

1 Answer
Oct 30, 2016

Answer:

# 4x^2-8x-12+6x = 4x^2-2x-12 = (4x+6)(x-2) #,
[ or #2(2x+3)(x-2)# ]

Explanation:

We want to factorise # 4x^2-8x-12+6x = 4x^2-2x-12 #

We look for two numbers which add to give the coefficient of #x#; so we seek two numbers which add to give #-2#.

However, as the coefficient of #x^2# is not #1# then instead of looking for two numbers which multiply to give the constant term #-12# we must look for two numbers which multiply to give #-48#, (that is, the coefficient of #x^2# multiplied by the constant term, ie #4 × −12#.

a × b = -48
a + b = -2

By inspection (or trial and error) we can find two numbers a=-8 and b=6

So we have,
# 4x^2-2x-12 = 4x^2-8x + 6x-12 #
# :. 4x^2-2x-12 = 4x(x-2) + 6x-12 # (by factorising the first two terms)
# :. 4x^2-2x-12 = 4x(x-2) + 6(x-2) # (collecting common terms)
# :. 4x^2-2x-12 = (4x+6)(x-2) # (by factorising the last two terms)