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

1 Answer
May 7, 2018

Answer:

#(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}#