Question

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

Answer 1

`(56+11x-16x^2)*10/(15x^2-11x-56)`

`= - {160x^2 -110 x - 560}/{15x^2 -11 x - 56}`

excluding `x = -8/5 and x=7/3`

Explanation:

There's a `15x^2` in the denominator and a `-16x^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\times 15= 3 \times 5`

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

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

`15x^2 -11 x - 56 = (5 x + 8) (3 x - 7)`

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

`x = -8/5 or x=7/3`

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

`(56+11x-16x^2)*10/(15x^2-11x-56)`

`= - {160x^2 -110 x - 560}/{15x^2 -11 x - 56}`