Polynomial Division 2: Step-by-Step Solution for (x^3-3x^2+12x-5)/(x-2)"

AI Thread Summary
The discussion focuses on performing polynomial division for the expression (x^3-3x^2+12x-5)/(x-2). The correct result of the division is shown to be (x^2-x+10) with a remainder of 15/(x-2). A mistake was noted in the initial attempt where the divisor was incorrectly written as 'x' instead of 'x - 2'. Participants emphasized the importance of verifying the solution by multiplying the quotient by the divisor and adding the remainder to check against the original polynomial. Overall, the conversation highlights the step-by-step process of polynomial division and the significance of accuracy in notation.
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Homework Statement



Show by polynomial division that

\frac{x^3-3x^2+12x-5}{x-2}=(x^2-x+10)+\frac{15}{x-2}

The Attempt at a Solution



Please see attachment
 

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In the attachment, you wrote the divisor as 'x' instead of 'x - 2' in the first line, which is incorrect.
 
so other than that my working is ok?
 
You can always check your work by multiplying the quotient by the divisor and adding any remainder, to see if you obtain the original dividend.
 
ah right that's what I have done on the bottom of the worksheets thanks :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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