How do you factor the expression #4x^2 + 16x + 15#?

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Answer 1 · 2015-12-30T18:30:50

#4x^2+16x+15=(2x+3)(2x+5)#

I'm going to explain this using the most common method of factorising: by splitting the middle term.

The first step is to multiply the coefficient of #x^2# with the constant. We get:

#4*15=60#

Now, we need to find the pair of factors of #60# whose sum or difference will give us the coefficient of #x#, i.e., #16#.

#60# has the following pairs of factors:

#(1,60), (2,30), (3,20), (5,12), (6,10)#

With a quick glance, it's clear that the sum of the factors in the pair #(6,10)# is #16#.

Great! So now we split the coefficient of middle term #(16)# as a sum of #6# and #10# as:

#4x^2 + (6+10)x + 15#
#4x^2 + 6x + 10x + 15#

Note: It doesn't matter if you reverse the order and split #16x# as #10x+6x#, you'll get the same result!

Now, we must take out common factors from the first two terms and then the next two terms:

#2x(2x+3)+5(2x+3)#

Now, we can take #(2x+3)# to be common, to get:

#(2x+3)(2x+5)#

and voila, that's the factored expression!