Question

Did I get the right answer for this question?? Multiplying and Dividing Rational Expressions.

Answer 1

This is the original problem.

Answer 2

This is what I got.

Answer 3

So the original question is \[{b^2+4b+4 \over 2b^2-8}\times{3b-6\over4b}\]is that right?

Answer 4

Yes :)

Answer 5

Ok well let's start by crossmultiplying the numerators by the denominators to get common factors.
\[{4b \times (b^2+4b+4\over4b(2b^2-8)}\times{(2b^2-8)(3b-6)\over4b(2b^2-8)}\]Notice how the bottoms are the same now so we can effectively write\[{4b \times (b^2+4b+4) \times(2b^2-8)(3b-6)\over4b(2b^2-8)}\]Does that make sense so far?

Answer 6

Yeah.

Answer 7

\[\cancel{4b} \times(b^2+4b+4) \times\cancel{(2b^2-8)}(3b-6) \over \cancel{4b(2b^2-8)}\]which leaves us with \[(b^2+4b+4) \times(3b-6)\] we can factorise the quadratic part and that will be the fully simplified answer

Answer 8

To factorise \[b^2+4b+4 \rightarrow (b+2)(b+2) \rightarrow (b+2)^2\]

Answer 9

So the final answer is \[(b+2)^2(3b-6)\] As a matter of style you could of course factorise out a 3 from the latter bracket.

Answer 10

I hope that helps!

Answer 11

Thank you :)