# How do you multiply (f^2 - 7f + 10)/( f^3 - 7f^2 + 7f - 49) *( f^2 - 49)/( 2f^3-10*f^2) ?

Sep 18, 2015

$\frac{{f}^{2} - 7 f + 10}{{f}^{3} - 7 {f}^{2} + 7 f - 49} \cdot \frac{{f}^{2} - 49}{2 {f}^{3} - 10 \cdot {f}^{2}} = \frac{\left(f - 2\right) \left(f + 7\right)}{\left({f}^{2} + 7\right) \left(2 {f}^{2}\right)}$

#### Explanation:

We know that ${f}^{2} - 49$ is a difference of squares which can also be written as $\left(f - 7\right) \left(f + 7\right)$

$2 {f}^{3} - 10 {f}^{2}$ has two factors we can put in evidence, $2$ and ${f}^{2}$, so we can rewrite it as $2 {f}^{2} \left(f - 5\right)$

${f}^{3} - 7 {f}^{2} + 7 f - 49$ is a bit more tricky, but you can get out by grouping by parts. To do this let's put ${f}^{2}$ in evidence whenever we can, so we have ${f}^{2} \left(f - 7\right) + 7 f - 49$

Now the two terms left have 7 as a common term, so we can put that in evidence and give us, ${f}^{2} \left(f - 7\right) + 7 \left(f - 7\right)$. Now we have $\left(f - 7\right)$ in both terms so we can put that in evidence, giving us the factored form: $\left({f}^{2} + 7\right) \left(f - 7\right)$

${f}^{2} - 7 f + 10$ is a quadratic so we can just use normal factoring methods for that:
Since the ${f}^{2}$ coefficient is 1 we know the factors will be of the form $\left(x + a\right) \left(x + b\right)$. Now, we know that $a b = 10$, and the two obvious names to come to mind are 2 and 5. We know that they must sum out to be the $f$ coefficient times $- 1$ so 2 and 5 fit. So now we put it all together and cancel everything that shows up on both numerator and denominator

$\frac{\left(f - 2\right) \cancel{\left(f - 5\right)}}{\left({f}^{2} + 7\right) \cancel{\left(f - 7\right)}} \cdot \frac{\cancel{\left(f - 7\right)} \left(f + 7\right)}{2 {f}^{2} \left(\cancel{f - 5}\right)} = \frac{\left(f - 2\right) \left(f + 7\right)}{\left({f}^{2} + 7\right) \left(2 {f}^{2}\right)}$

Factoring functions like this can be a bit hard, specially for non-quadratics, but I've found that this is a good source..