How do you factor the trinomial #2x^2 + 9x + 4#?

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Answer 1 · 2016-01-03T14:44:30

#2x^2+9x+4=(2x+1)(x+4)#

#2x^2+9x+4#
First multiply the constant #4# with the coefficient of #x^# which is #2#

The result is #8#

Now find factors of #8# such that they add up to the coefficient of #x# here it is #+9#.

Let us list out the pairs which multiply to give #8#
These are #1,8#, #2,4#, #-1,-8# and #-2,-4#

Let us check if these add up to #+9#

#1xx8=8 and 1+8 =9# this works. Using this we shall split the middle term. #9x = 1x+8x#

#2x^2+9x+4#
#=2x^2+1x+8x+4# This is called splitting the middle term.
Now we would do factor by grouping.

#=(2x^2+1x)+(8x+4)#

GCF from each group is taken out.

#=x(2x+1) + 4(2x+1)#

Now we have #(2x+1)# as the common factor, so it is factored out.

#=(2x+1)(x+4)# Answer