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This is in an IGCSE additional math question relating to factor and remainder theorem.

My attempt:

Using factor theorem we know that $f(1)=0$, $f(k)=0$ and $f(k+1)=0$. Using remainder theorem we know that $f(2)=20$.

From this we can form the following equations (unless I made a careless mistake):

$$a+b+c=-1$$ $$4a+2b+c=-8$$ $$k^3+ak^2+bk+c=0$$ $$k^3+3k^2+ak^2+3k+2ak+b+1=0$$

As the last two equations are both equal to zero, we can form the equation:

$$3k^2+3k+2ak+b+1=0$$

This is as far as I was able to come, I'd hoped that I would find a trick to 'solve' the question at this point but haven't been able to come up with anything. So I'm looking to be pointed in the right direction from here (or a solution).

Thanks.

P.S. This is not a homework or assignment, I'm just self-studying a lot of the topics I'll be seeing in A level math a few months from now. (Also my first question here.)

CroW
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2 Answers2

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Since you know the roots of $f(x)$ you have

$$f(x) = (x-1)(x-k)(x-k-1).$$

So

$$f(2) = (2-1)(2-k)(2-k-1) = k^2-3k+2 = 20,$$

which is to say $k^2-3k-18 = 0.$

B. Goddard
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  • I can't believe the solution was this simple all along! Thanks a lot for your time. – CroW Jun 01 '22 at 17:06
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Note: I think your second equation should be $4a+2b+c=12$ (which will be crucial to solving the problem later) and $k^3+3k^2+ak^2+3k+2ak+bk+1+a+b+c=0$ instead of $k^3+3k^2+ak^2+3k+2ak+b+1=0$.
From Vieta’s formulas, $$-a=1+k+(k+1)=2(k+1)$$ $$b=k.1+(k+1).1+k.(k+1)=k^2+3k+1$$ $$-c=1.k.(k+1)=k^2+k.$$ Substituting these into $4a+2b+c=12$ gives the desired expression.

insipidintegrator
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  • Thanks for the answer. The first equation was a blunder on my part but the second equation was just a mistype in the question submitted here. I also didn't know about the Vieta's formulas but I'll certainly look into them soon now that I know about them! – CroW Jun 01 '22 at 17:05