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Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$?

The question is mentioned in my math olympiad. Please explain how to solve the problem. I have factorised the equation to $$-x^2+1, x-6x, -5x +1.$$

I am only in year 6.

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The expression factors as $(x-r)(x-s)(x-t)$ so that your desired expression is just the original polynomial evaluated at $2$. The answer is $-5$. –  Jared Apr 5 '13 at 1:16

5 Answers 5

Your polynomial is monic (leading coefficient is $1$), of degree $3$, and you are given three roots. You have no choice (since the coefficients here are in a field, say $\mathbb{R}$), it is completely factored as follows: $$ p(x)=x^3-6x^2+5x+1=(x-r)(x-s)(x-t). $$ Now what does it take to compute the value of your expression?

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I left the OP something to do. –  amWhy Apr 5 '13 at 2:29
    
I highly doubt that a year six student would be familiar with fields... –  manthanomen Apr 5 '13 at 3:44
    
@user67750 I know, but I couldn't help it... But how do you explain a year six student the factorization? I guess you just hide it under the rug. –  1015 Apr 5 '13 at 3:48

If $r,s,t$ be the roots of the equation $x^3 - 6x^2 + 5x + 1=0$, then $$x^3 - 6x^2 + 5x + 1 = (x - r)(x - s)(x - t)$$

Now, determine the value of $f(2)$ by evaluating $$x^3 - 6x^2 + 5x + 1\quad\text{at} \;x = 2\;\tag{$f(2)$}$$ Then set $$(2 - r)(2-s)(2 - t) = f(2)\tag{1}$$

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And I was trying to leave the OP something to do... –  1015 Apr 5 '13 at 1:22

Note that $$x^3-6x^2+5x+1=(x-r)(x-s)(x-t).$$ Does the right-hand side look familiar (perhaps if we replaced $x$ with something)?

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What are you trying to find the value of? Do you see the similarity in form? –  Cameron Buie Apr 5 '13 at 1:23

If $c$ is a root of some polynomial, then $x - c$ is a a factor of that polynomial. Furthermore, if the polynomial has degree $n$, then it has at most $n$ roots. So $$x^3 - 6x^2 + 5x + 1 = (x - r)(x - s)(x - t)$$ Thus, \begin{align*} (2 - r)(2 - s)(2 - t) &= 2^3 -6(2^2) + 5(2) + 1 \\ &= 8 - 6(4) + 10 + 1 \\ &= 8 - 24 + 10 + 1 \\ &= -5 \end{align*}

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Hint $\ $ If $\rm\:r,s,t\:$ are roots of $\rm\:f(x)\:$ then $\rm\:2\!-\!r,2\!-\!s,2\!-\!t\:$ are roots of $\rm\:f(2\!-\!x),\:$ so their product is minus the constant term of $\rm\:-f(2\!-\!x),\:$ i.e. $\rm\:f(2) = -5\ $ (we use $\rm\ {-}f\:$ so it has lead coef $= 1).$

Remark $\ $ Similarly $\rm\:(2\!-\!r)(2\!-\!s)+(2\!-\!s)(2\!-\!t)+(2\!-\!t)(2\!-\!r) =\:$ coef of $\rm\:x\:$ in $\rm\:g(x) = -f(2\!-\!x),\:$ i.e. $\rm\ g'(0) = f'(2) = -7.$

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