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

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The discussion revolves around factoring a polynomial denominator without dividing by the highest common factor (H.C.F.). The numerator is confirmed to be x(x+4)(x-1), with x-1 identified as a factor of the denominator. The denominator, 7x^3-18x^2+6x+5, is manipulated by splitting -18x^2 into -7x^2 and -11x^2, allowing for further factoring. Additional steps involve rewriting 6x as -11x - 5x to facilitate factoring two terms at a time. The participants express understanding and appreciation for the clarification provided.
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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.?
 

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Miike012 said:
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.?
attachment.php?attachmentid=37714&d=1312339305.jpg
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.
 


Got it.. thank you.
 

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