How do you factor #2x^2-21x-11#?

1 Answer
Oct 30, 2016

Answer:

#2x^2-21x-11 = (x-11)(2x+1) #

Explanation:

We want to factorise #2x^2-21x-11#

We look for two numbers which add to give the coefficient of #x#; so we seek two numbers which add to give #-21#.

However, as the coefficient of #x^2# is not #1# then instead of looking for two numbers which multiply to give #−11# we must look for two numbers which multiply to give #−22#, (that is, the coefficient of #x^2# multiplied by the constant term, #2 × −11#.

a × b = -22
a + b = -21

By inspection (or trial ad error) we can find two numbers a=-22 and b=1

So we have,
# 2x^2-21x-11 = 2x^2-22x + x-11 #
# :. 2x^2-21x-11 = 2x(x-11) + x-11 # (by factorising the first two terms)
# :. 2x^2-21x-11 = 2x(x-11) + (x-11) # (collecting common terms)
# :. 2x^2-21x-11 = (x-11)(2x+1) # (by factorising the last two terms)

We can check this by multiplying out:
# (x-11)(2x+1) = 2x^2 + x -22 x -11 = 2x^2-21x-11 # QED