How do you factor the expression #-4x^2 + 10x + 24#?

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Answer 1 · 2015-12-30T19:33:05

#(4-x)(4x+6)#

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*24=-96#

Now, we need to find the pair of factors of #-96# whose sum or difference will give us the coefficient of #x#, i.e., #10#.

#-96# has the following pairs of factors:

#(96,-1), (32,-3), (16,-6), (8,-12), (4,-24)# as well as all these pairs with a reversal of signs.

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

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

#-4x^2 + (16-6)x + 24#
#-4x^2 + 16x - 6x + 24#

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

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

#4x(-x+4)+6(-x+4)#

Now, we can take #(-x+4)# to be common, to get:

#(-x+4)(4x+6)#

or, #(4-x)(4x+6)#

and voila, that's the factored expression!