How do you multiply #(6x^2-54x)/(x-9)*(7x)/(6x)# and state the excluded values?

1 Answer
Oct 23, 2017

Answer:

Multiply numerators together to find new numerator, and multiply denominators to find new denominator. 9 and 0 are excluded, as each one sets a denominator equal to 0.

#7x#

Explanation:

The excluded values will be those wherein the denominator is 0. This is the case for any situation where either denominator of the factors would be 0. In this case, those would be #x=9# (which would make #x-9=0#), and #x=0#.

We multiply these expressions by multiplying the numerators and denominators separately. The product of the numerators will be the new numerator, and the product of the denominators will be the new denominator.

#(6x^2-54x)/(x-9) * (7x)/(6x) = (7x(6x^2-54x))/(6x(x-9)) = (42x^3-378x^2)/(6x^2 - 54x) #

We can factor out both an #x# and a #6# (378 was the product of 54 and 7, and 6 is a factor of 54, thus it is a factor of 378). We can only do this factoring if #x!=0#, as otherwise the denominator for our original product becomes #6(0)^2 -54(0) = 0#, and we cannot divide by 0.

#= (7x^2-63x)/(x-9)#

Note that now the numerator is a multiple of the denominator...

#= (7x(x-9))/(x-9)#

And for any #x!=9# (since otherwise our denominator, once more, becomes 0...

#=7x#