How do you simplify and find the excluded values of #(x^2+5x+4)/(x^2-x-72)#?

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Answer 1 · 2016-06-20T07:02:04

the exclude values are #x=9 and x=-8#

Try the factoring of both trynomials by using the following rule:

in a trynomial #x^2+sx+p# find two numbers a and b of which s is the sum and p the product, then factor in this way:

#(x+a)(x+b)#

so

#x^2+5x+4=(x+1)(x+4)#
and
#x^2-x-72=(x-9)(x+8)#

The fraction is equivalent to the following one:

#((x+1)(x+4))/((x-9)(x+8))#

but you cannot symplify the expression!

The excluded values are those ones that make null the denominator:

#(x-9)(x+8)#

that are #x=9 and x=-8#