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If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:

I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial.

I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.

Does a better method than the lackluster substitution, exist?

The answer is:

$x^2-3x+2$

3 Answers3

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Hint:

As $a,b$ are the roots of $x^2-2x+3=0$

$a^3-3a^2+5a-3=(a^2-2a+3)(a-1)+1=1$

Similarly for $b$

lab bhattacharjee
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Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then, $$ \begin{align} a^3-3a^2+5a-2 &=(a-6)-3(2a-3)+5a-2\\ &=1 \end{align} $$ and $$ \begin{align} b^3-b^2+b+5 &=(b-6)-(2b-3)+b+5\\ &=2 \end{align} $$ It is easy to find an equation which has roots of $1$ and $2$.

robjohn
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  • Never thought about using the initial polynomial to create a reduction formula absolutely radical! –  Dec 24 '18 at 16:43
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    @CaptainQuestion In this case it's actually easier to use long division to compute the remainder - I added that to my answer to show how easy it is. – Bill Dubuque Dec 24 '18 at 16:44
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Hint $\ x^3-3x^2+5x-2\,\bmod\, \color{#c00}{x^2-2x+3}\, =\, \color{#0a0}1\ $ (and $= \color{#90f}2$ for the other). So we seek a polynomial with roots $\color{#0a0}1$ and $\color{#90f}2,\,$ e.g. $\ (x-\color{#0a0}1)(x-\color{#90f}2)$

Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)

$$\begin{align} &\ \ \ 1\ {-}3\ \ \ \, 5\,\ {-}2\\ &\color{#c00}{{-}1\,\ \ \ 2\ {-}3}\\ & \ \ \ \ \ \ {-}1\ \ \ \ 2\ \ {-}2\\ &\color{#c00}{\ \ \ \ \ \ \ \ \ 1\ {-}2\ \ \ \ \ 3}\\ &\qquad\qquad\quad\ \ \color{#0a0}1 \end{align}\qquad\qquad\quad$$

Bill Dubuque
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