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- Feb 14, 2012

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Let $P(x)=x^3-2x+1$ and $Q(x)=x^3-4x^2+4x-1$. Show that if $P(r)=0$, then $Q(r^2)=0$.

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- Thread starter
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- #1

- Feb 14, 2012

- 3,937

-----

Let $P(x)=x^3-2x+1$ and $Q(x)=x^3-4x^2+4x-1$. Show that if $P(r)=0$, then $Q(r^2)=0$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

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1. kaliprasad

2. lfdahl

Solution from lfdahl :

\[0 = [P(r)]^2 = \left [ r^3-2r+1 \right ]^2 \\\\ = r^6-4r^4+2r^3+4r^2-4r+1 \\\\ = \underbrace{r^6 - 4r^4 + 4r^2-1}_{= Q(r^2)} +\underbrace{2r^3-4r + 2}_{= 2P(r)} = Q(r^2)\;\;\;\; q.e.d.\]

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