Question 629403: ((3a^2-13a+4)/(9a^2-6a+1))((28+7a)/(a^2-16))
Answer by perfectdose(7)   (Show Source):
You can put this solution on YOUR website! Each section has to be factored seperatly.
First I'll do the numerator of the first fraction
3a^2-13a+4, we have to find factors of A (times)* C that adds up to B.
3=a -13=b 4=c
multiple a*c, 3*4=12 that adds up to -13
-12*-1=12 and -12+-1=-13, so those are my two factors.
now we replace the middle term with the two factors.
3a^2-13a+4 the -13a gets replaces with the -12a and -1a (matching A's)
(3a^2-12a)+(-1a+4) now we have four terms so we group and factor out what we can from each group.
3a(a-4)-1(a-4)
Take inner and outter terms (a-4)(3a-1)
Second I'll do the denominator in the first fracton.
Doing the same thing our factors that multiple to a 9 and add to a -6 = -3*-3=9 and -3+-3=-6
replace the middle term with the two terms and factor what you can out from each group.
(9a^2-3a)+(-3a+1)
3a(3a-1)-1(3a-1) take inner and outer terms.
(3a-1)(3a-1)
Now I'll do the numerator of the second fraction
28+7a its only two terms so we see if theres a GFC (a number that goes into both numbers)
7(4+a)
Second I'll do the denominator in the second fracton
a^2-16
is a perfect square.
(a+4) (a-4)
so we have
(a-4)(3a-1)/(3a-1)(3a-1) * 7(4+a)/(a+4) (a-4)
next we cancel out terms that match in the top and bottom and we're left with
7/(3a-1)
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