problem to Brilliant ! 5

Algebra Level 5

If p , q & r are the roots of polynomial x³+4x²-3x+5 =0 Find the value of (p⁵+q⁵+r⁵) + (p⁴+q⁴+r⁴) – (p²+q²+r²)²


The answer is -2577.

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1 solution

Chew-Seong Cheong
Dec 23, 2014

This problem can be solved using Newton's Sums method.

Let S 1 = p + q + r = 4 S_1 = p+q+r = -4 , S 2 = p q + q r + r p = 3 \space S_2 = pq+qr+rp = -3 , S 3 = p q r = 5 \space S_3 = pqr = -5\space and P n = p n + q n + r n P_n = p^n + q^n + r^n , where n = 1 , 2 , 3 n = 1,2,3 .... Then P n P_n are given by:

{ P 1 = S 1 = 4 P 2 = S 1 P 1 2 S 2 = 4 ( 4 ) 2 ( 3 ) = 22 P 3 = S 1 P 2 S 2 P 1 + 3 S 3 = 4 ( 22 ) + 3 ( 4 ) + 3 ( 5 ) = 155 P 4 = S 1 P 3 S 2 P 2 + S 3 P 1 = 4 ( 155 ) + 3 ( 22 ) 5 ( 4 ) = 546 P 5 = S 1 P 4 S 2 P 3 + S 3 P 2 = 4 ( 546 ) + 3 ( 155 ) 5 ( 22 ) = 2639 \begin {cases} P_1 = S_1 & & = -4 \\ P_2 = S_1 P_1 - 2S_2 & = -4(-4) - 2(-3) &= 22 \\ P_3 = S_1 P_2 - S_2P_1 + 3S_3 & = -4(22) + 3(-4) + 3(-5) &= -155 \\ P_4 = S_1 P_3 - S_2P_2 + S_3P_1 & = -4(-155) + 3(22) -5(-4) &= 546 \\ P_5 = S_1 P_4 - S_2P_3 + S_3P_2 & = -4(546) + 3(-155) -5(-22) &= -2639 \end {cases}

Therefore,

( p 5 + q 5 + r 5 ) + ( p 4 + q 4 + r 4 ) ( p 2 + q 2 + r 2 ) 2 (p^5+q^5+r^5) + (p^4+q^4+r^4) - (p^2+q^2+r^2)^2

= P 5 + P 4 P 2 2 = 2639 + 546 ( 2 2 2 ) = 2577 = P_5+P_4-P_2^2 = -2639+546-(22^2) = \boxed {-2577}

Sir please tell me that is there any formula to calculate P n P_n

means how can we calculate P 8 P_8 or P 9 P_9 With respect To S n S_n

Mehul Chaturvedi - 6 years, 5 months ago

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Mehul, for a k t h k^{th} degree polynomial with roots a 1 , a 2 , a 3 . . . a k a_1, a_2, a_3... a_k , there will be S 1 , S 2 , S 3 . . . S k S_1, S_2, S_3... S_k . This question is for 3 3 roots therefore there are S 1 , S 2 , S 3 S_1, S_2, S_3 . Then, P n P_n is given by:

P n = { S 1 P n 1 S 2 P n 2 + S 3 P n 3 . . . + ( 1 ) n 1 n S n n k S 1 P n 1 S 2 P n 2 + S 3 P n 3 . . . + ( 1 ) k 1 S k P n k n > k P_n = \begin {cases} S_1P_{n-1}-S_2P_{n-2}+S_3P_{n-3}...+(-1)^{n-1}nS_n & n \le k\\ S_1P_{n-1}-S_2P_{n-2}+S_3P_{n-3}...+(-1)^{k-1}S_kP_{n-k} & n > k \end {cases}

For this question:

P 6 = 4 ( P 5 ) + 3 ( P 4 ) 5 ( P 3 ) = 4 ( 569 ) + 3 ( 162 ) 5 ( 42 ) = 2005 P_6 = -4(P_5)+3(P_4)-5(P_3) = -4(-569)+3(162)-5(-42) = 2005

P 7 = 4 ( 2005 ) + 3 ( 569 ) 5 ( 162 ) = 7123 P_7 = -4(2005)+3(-569)-5(-162) = -7123

P 8 = 4 ( 7123 ) + 3 ( 2005 ) 5 ( 569 ) = 25322 P_8 = -4(-7123)+3(2005)-5(-569) = 25322

P 9 = 4 ( 25322 ) + 3 ( 7123 ) 5 ( 2005 ) = 89944 P_9 = -4(25322) +3(-7123)-5(2005) = -89944

Chew-Seong Cheong - 6 years, 5 months ago

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A big thanks to you sir!!

Mehul Chaturvedi - 6 years, 5 months ago

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@Mehul Chaturvedi You don't even need to remember the formula. Just plug in the roots (in the polynomial) in succession and add, you'll get the same result.

You can also multiply the polynomial by a power of x x , since this won't change anything (Although you are introducing more roots, but that doesn't matter) and then plug in the roots and then add to get the sum.

Aneesh Kundu - 6 years, 5 months ago

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@Aneesh Kundu Would you please clarify more?

Mehul Chaturvedi - 6 years, 5 months ago

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@Mehul Chaturvedi For simplicity lets just assume that there are 3 roots(although this can be generalised easily). Let S n S_n denote the sum of the n th n^{\text{th}} powers of the roots, i.e. S n = α n + β n + γ n S_n=\alpha^n+\beta^n+\gamma^n let α , β , γ \alpha, \beta, \gamma be the roots of f ( x ) = a x 3 + b 2 + c x + d = 0 f(x)=ax^3+b^2+cx+d=0 Since α , β , γ \alpha, \beta, \gamma are the roots of f ( x ) = 0 f(x)=0 , f ( α ) = a α 3 + b α 2 + c α + d = 0 f(\alpha)=a\alpha^3+b\alpha^2+c\alpha+d=0 f ( β ) = a β 3 + b β 2 + c β + d = 0 f(\beta)=a\beta^3+b\beta^2+c\beta+d=0 f ( γ ) = a γ 3 + b γ 2 + c γ + d = 0 f(\gamma)=a\gamma^3+b\gamma^2+c\gamma+d=0 Adding all the eqns, we get a S 3 + b S 2 + c S 1 + 3 d = 0 aS_3+bS_2+cS_1+3d=0 What I did above can also be stated as plugging the roots in succession and then adding.

Well, if we know S 2 S_2 and S 1 S_1 we can easily calculate S 3 S_3 . But we could calulate S 3 S_3 even without that, so how is that useful? In the next step we draw some useful results and try to generalize a bit. We started with f ( x ) = a x 3 + b 2 + c x + d = 0 f(x)=ax^3+b^2+cx+d=0 what if multiply both sides by x n 3 x^{n-3} ? Lets just try that a x n + b n 1 + c x n 2 + d x n 3 = 0 ax^n+b^{n-1}+cx^{n-2}+dx^{n-3}=0 Since we just multiplied the equation by a number, this still hasn't affected anything(although we have introduced more roots), this means that α , β , γ \alpha, \beta ,\gamma are still its roots.

Now we plug-in the roots in succession and add.

a S n + b S n 1 + c S n 2 + d S n 3 = 0 aS_{n}+bS_{n-1}+cS_{n-2}+dS_{n-3}=0

This is a very useful result and it can help us in calculating S n S_{n} if we know S n 1 S_{n-1} , S n 2 S_{n-2} , S n 3 S_{n-3} (and also the coefficients of the polynomial).

Hope this helped.

Aneesh Kundu - 6 years, 5 months ago

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@Aneesh Kundu Thanks for your explanation. I will try it out in the next question.

Chew-Seong Cheong - 6 years, 5 months ago

From which book,You read this?

Mehul Chaturvedi - 6 years, 5 months ago

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@Mehul Chaturvedi I googled and did some thinking. http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums. Read also Vieta's formulas.

Chew-Seong Cheong - 6 years, 5 months ago

I did same

Dev Sharma - 5 years, 7 months ago

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