How do you factor #6p^2 - p - 15#?

1 Answer
Joel E.
Mar 3, 2018

#(2p+3)(3p-5)#

Explanation:

#6p^2-p-15#
if it helps, think about it like this: #ap^2-bp-c#

First, multiply "a" and "c"
#6*-15=-90#

Find 2 factors of -90 that give you -1 (or the b-value) when you add them

1,-90 doesn't work because it adds to -89
2, -45 doesn't work because it adds to -43
3, -30 doesn't work because it adds to -27
5, -18 doesn't work because it adds to -13
6, -15 doesn't work because it adds to -9
9,-10 Works because it adds to #-1# !!

Now that you got the two factors of -90 that add up to -1, set it all up in this format: #(p+9/a)(p-10/a)#

so you'll get:
#(p+9/6)(p-10/6)#

simplify
#(p+3/2)(p-5/3)#

if you can't simplify anymore, take the denominator and make it p's coefficient and now you are finished
#(2p+3)(3p-5)#

If you're still stuck, use these helpful links:

math is fun

khan academy

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