Ive added an attachment and highlited the area that I have a question

[Miike012]

Ive added an attachment and highlited the area that I have a question about. How where they able to resolve the denominator into factors without dividing the denominator by the H.C.F.?

 

[SammyS]

Ive added an attachment and highlighted the area that I have a question about. How where they able to resolve the denominator into factors without dividing the denominator by the H.C.F.?

Once it's determined that the numerator in factored form is, x(x+4)(x-1), it's easy to see,by using the remainder theorem, that the only one of these that's a factor of the denominator is x-1.

The denominator is 7x^3-18x^2+6x+5\,.

Split up -18x² into -7x² - 11x²

You get 7x^3-7x^2-11x^2+6x+5\quad\to\quad7x^2(x-1)-11x^2+6x+5 \,.

Similarly write 6x as -11x - 5x & factor as needed two terms at a time.

 

[Miike012]

Got it.. thank you.