Question

I am having a mental block on this question! Any help?

Answer 1

\[\frac{a ^{4} }{2a ^{2}-4a ^{2} } \times \frac{8a ^{3}+1}{ 2a+4a ^{2} }\]

Answer 2

Just multiply the top witht the top and the bottom with the bottom, can you proceed formt here or is that not the problem?

Answer 3

Are you sure you wrote the denominator correctly in the second fraction? My suspicion is that perhaps it should be \(2a^2+4a^2\)

I would also factor what you have before multiplying, in the hopes of canceling any common factors first.

Answer 4

Nope whpalmer4 ... it is just 2a

Answer 5

\[\frac{a ^{4} }{2a ^{2}-4a ^{2} } \times \frac{8a ^{3}+1}{ 2a+4a ^{2} }\]
\[\frac{a ^{4} }{a ^{2}(2-4) } \times \frac{8a ^{3}+1}{ a(2+4a) }\]
\[\frac{a}{2-4} \times \frac{8a ^{3}+1}{ 2+4a}\]
\[\frac{a}{-2} \times \frac{8a ^{3}+1}{ 2+4a}\]

Answer 6

Hold on there, you aren't done yet :-)

Answer 7

That can still be factored and simplified further.

\[-\frac{a}{2}*\frac{8a^3+1}{2(2a+1)} = -\frac{a}{4}*\frac{(2a+1)(4a^2-2a+1)}{(2a+1)} =\]\]

Answer 8

I couldn't be expected to do ALL the work.. :P

Answer 9

oh, and I didn't either, just enough to show that there was more work to be done ;-)

I didn't notice when the question was actually closed, perhaps the OP is long gone.

Answer 10

I think it was the OP was here earlier.

Answer 11

Thanks guys... I made a silly mistake in the second step!