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OpenStudy (j2lie):
How to factor 18a^2+15a-18?
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OpenStudy (j2lie):
@whpalmer4 can you help me?
OpenStudy (whpalmer4):
first, do you see any common factors in the coefficients?
OpenStudy (j2lie):
yes 3.
OpenStudy (whpalmer4):
okay, generally it makes life easier to factor out any common factors first, so the numbers we work with are as small as possible
OpenStudy (whpalmer4):
so that gives us \[3(6a^2+5a-6)\]right?
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OpenStudy (j2lie):
yes
OpenStudy (whpalmer4):
do you remember the procedure for splitting the middle when the leading term has a coefficient other than 1?
OpenStudy (whpalmer4):
we'll just leave the 3 behind for the moment and factor
\[6a^2+5a-6\]
OpenStudy (j2lie):
6a^2+6a-1a_6
OpenStudy (whpalmer4):
huh?
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OpenStudy (whpalmer4):
so the procedure is to multiply the first and last term's coefficients:
6*-6 = -36
Now we look for a pair of factors of -36 that sum to +5
OpenStudy (j2lie):
you have to split the middle coefficient.
OpenStudy (j2lie):
this is my answer #(2a+3)(3a-2)
OpenStudy (j2lie):
3*
OpenStudy (whpalmer4):
that's correct!
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OpenStudy (anonymous):
3(2x+3)(3x−2)
OpenStudy (j2lie):
yay my dad said I'm wrong.
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