How do you factor completely #2x^2+6x-80#?

2 Answers
May 12, 2016

Answer:

y = 2(x - 5)( x + 8)

Explanation:

#y = 2(x^2 + 3x - 40)#
Factor the trinomial in parentheses.
Find 2 numbers knowing sum (b = 3) and product (c = -40).
They are (-5, 8).
y = 2(x - 5)(x + 8)

May 12, 2016

Answer:

#2x^2+6x-80=color(blue)(2(x-5)(x+8))#

Explanation:

Given
#color(white)("XXX")2x^2+6x-80#

First observe that all the terms have a constant factor of #2#;
so we can write:
#color(white)("XXX")2(x^2+3x-40)#

Next lets think about factors of #40# whose difference is #3#

#40=2xx20# ...not a difference of #3#
#40=4xx10# ...not a difference of #3#
#40=5xx8# ...looks as if we've found it

In #(x^2+3x-40)# the last term #(-40)# is negative so one of #5# or #8# must be negative;
and the middle term #(+3x)# is positive so the larger of #5# and #8# must be positive.

Therefore our complete factors are:
#color(white)("XXX")2(x-5)(x+8)#