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HACTEHA [7]
4 years ago
13

Please help! Divide 3a^2+2a-1 by a^2-1

Mathematics
1 answer:
Ludmilka [50]4 years ago
5 0
So one way is to factor and cross out zeroes

so factor 3a^2+2a-1
write the factors
(3a  )(a )
now write the end factors
(3a-1)(a+1)
mutiply
(3a^2-a+3a-a)=3a^2+2a-1 correct

factor a^2-1
difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)
a^2-1^2=(a-1)(a+1)

so now we have
\frac{3a^{2}+2a-1}{a^2-1}= \frac{(3a-1)(a+1)}{(a-1)(a+1)}  = (\frac{(3a-1)}{(a-1)})  (\frac{(a+1)}{(a+1)}) = (\frac{(3a-1)}{(a-1)}) (1)= \frac{(3a-1)}{(a-1)}

the answer is \frac{(3a-1)}{(a-1)}
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