# How do you multiply (56+11x-16x^2)*10/(15x^2-11x-56) and state the excluded values?

May 7, 2018

$\left(56 + 11 x - 16 {x}^{2}\right) \cdot \frac{10}{15 {x}^{2} - 11 x - 56}$

$= - \frac{160 {x}^{2} - 110 x - 560}{15 {x}^{2} - 11 x - 56}$

excluding $x = - \frac{8}{5} \mathmr{and} x = \frac{7}{3}$

#### Explanation:

There's a $15 {x}^{2}$ in the denominator and a $- 16 {x}^{2}$ in the numerator, so the cancelling that we might hope for at first glance doesn't materialize.

The excluded values occur at the zeros of the denominator; let's try to factor. We seek a pair of factors of $15$ and a pair of factors of $- 56$ whose sum of products is $- 11.$ That's a bit of a search, we have

$15 = 1 \setminus \times 15 = 3 \setminus \times 5$

$56 = 1 \setminus \times 56 = 2 \setminus \times 28 = 4 \setminus \times 14 = 7 \setminus \times 8$ with a minus sign in there for $- 56.$

Eventually we find $- 7 \setminus \times 8 = - 56 ,$ $5 \times 3 = 15 ,$ -7(5)+3(8)=-11 quad sqrt

$15 {x}^{2} - 11 x - 56 = \left(5 x + 8\right) \left(3 x - 7\right)$

We could also have used the quadratic formula to find the zeros of the denominator, which are called poles.

$x = - \frac{8}{5} \mathmr{and} x = \frac{7}{3}$

The multiplication itself is rather vacuous due to the lack of cancelling.

$\left(56 + 11 x - 16 {x}^{2}\right) \cdot \frac{10}{15 {x}^{2} - 11 x - 56}$

$= - \frac{160 {x}^{2} - 110 x - 560}{15 {x}^{2} - 11 x - 56}$