How do you factor completely #x^2+5x+7x+35#?

1 Answer
Nov 12, 2015

Answer:

Factor by grouping to find:

#x^2+5x+7x+35 = (x+7)(x+5)#

Explanation:

#x^2+5x+7x+35 = (x^2+5x)+(7x+35) = x(x+5)+7(x+5)#

#= (x+7)(x+5)#

With this problem, the hard work has already been done for you in the way that the middle terms are split as #5x+7x#. Normally you would be faced with something like #x^2+12x+35# and have to find the split. The trick is to find two numbers which add up to the coefficient #12# of the term in #x# and multiply together to give the constant term #35#.

Here's another example:

#x^2+14x+40#

We want to find two numbers that add up to #14# and multiply to give #40#. Having found that #10# and #4# work, we have:

#x^2+14x+40 = (x+10)(x+4)#

In general, we can say:

#(x+a)(x+b) = x^2+(a+b)x+ab#