Question

How do you simplify `(x^2+6x-27)div(3x^2+27x)/(x+5)`?

Answer 1

If `x!=-9` and `x!=3` then

`(x^2+6x-27)div(3x^2+27x)/(x+5)=color(green)((x^2+2x-15)/(3x))`

Explanation:

Note that
`color(white)("XXX")x^2+6x-27=(x+9)(x-3)`
and that
`color(white)("XXX")3x^x+27=3x(x+9)`

and since `q div a/b rArr q * b/a`

`(x^2+6x-27) div (3x^2+27x)/(x+5)`

`color(white)("XXX")=(cancel(x+9))(x-3)xx(x+5)/(3x(cancel(x+9)))color(white)("XXX")`
`color(white)("XXX")`provided `x+9!=0 and 3x!=0`

`color(white)("XXX")=(x^2+2x-15)/(3x)`

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Note that if `x=-9` or `x=3`
then `3x^2+27x=0` and the division is undefined