How do you factor #4p ^ { 2} - 16p - 84#?

2 Answers
Jul 18, 2017

#4p^2-16p-84#

=#4p^2-16p+16-100#

=#(2p-4)^2-10^2#

=#(2p-4+10)*(2p-4-10)#

=#(2p+6)*(2p-14)#

Explanation:

I used difference of squares identity.

Jul 18, 2017

#4p^2-16p-84 = 4(p-7)(p+3)#

Explanation:

Given:

#4p^2-16p-84#

Note that all of the coefficients are divisible by #4#, so we can separate that out as a constant factor.

The remaining quadratic is monic (leading coefficient #1#), so we then just need to find two factors of #21# which differ by #4#. The pair #7, 3# works, so we find:

#4p^2-16p-84 = 4(p^2-4p-21)#

#color(white)(4p^2-16p-84) = 4(p-7)(p+3)#