How do you factor #5x^{2}+18x-35#?

1 Answer
Oct 14, 2016

The factorization of #5x^2 + 18x - 35# is #(x + 5)(5x - 7)#.

Explanation:

Remember that #5x^2# is the result of "first times first" multiplication, #18x# is the sum of "outside times outside" and "inside times inside" multiplications, and #-35# is the result of "last times last" multiplication.

To begin, think about the factor pairs whose product is #35#.
#1*35# and #5*7# are the only options.

Now consider the factor pair whose product is #35x^2#.
#1x*5x# is the only one.

We know that the two "first" terms in the factorization will be #1x# and #5x#. We can easily discard #1*35# as the factor pair to obtain the product of #35#, because of the large sums that would result from adding the outside & inside products.

Therefore we must decide on the proper placement of the #5# and #7#, along with one #+# and one #-# & the effect of the #5x#, to get the outside & inside sum of #18x#.

#(x + 7)(5x - 5)# does not work, because the "outside times outside" multiplication results in #-5x# & the "inside times inside" multiplication results in #35x#. The sum of these two products is #30x#. Try again.

#(x + 5)(5x - 7)# works. Do FOIL to verify:

#(x + 5)(5x - 7)#
#5x^2 - 7x + 25x - 35#
#5x^2 + 18x - 35#

This verifies that the correct factorization is
#(x + 5)(5x - 7)#.