How do you factor #18c^2-3c-10#?

1 Answer
Apr 16, 2016

Answer:

#18c^2-3c-10=(6c-5)(3c+2)#

Explanation:

We can use an AC method:

Look for a pair of factors of #AC = 18*10 = 180#, with difference #B=3#.

The pair #12, 15# works.

Then use this pair to split the middle term and factor by grouping:

#18c^2-3c-10#

#=18c^2+12c-15c-10#

#=(18c^2+12c)-(15c+10)#

#=6c(3c+2)-5(3c+2)#

#=(6c-5)(3c+2)#

#color(white)()#
Alternative method

Alternatively, we can find this factorisation by completing the square then using the difference of squares identity:

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

with #a=(12c-1)# and #b=9# as follows:

To make the arithmetic easier, first multiply by #8#, remembering to divide by #8# at the end:

#8(18c^2-3c-10)#

#=144c^2-24c-80#

#=(12c^2)-2(12c)-80#

#=(12c-1)^2-1-80#

#=(12c-1)^2-81#

#=(12c-1)^2-9^2#

#=((12c-1)-9)((12c-1)+9)#

#=(12c-10)(12c+8)#

#=(2(6c-5))(4(3c+2))#

#=8(6c-5)(3c+2)#

So dividing both ends by #8# we find:

#18c^2-3c-10 = (6c-5)(3c+2)#

Why did I multiply by #8# first?

If I multiplied by #2# then that would make the leading term #36c^2 = (6c)^2#, but then the completed square would have involved #(6c-1/2)^2 = 36c^2-6c+1/4#. To avoid the messiness of fractions, multiply by an extra factor of #2^2 = 4# first.