Polynomial Roots - 1

Algebra Level 4

Let p p , q q and r r be distinct real roots of the equation x 3 4 x 2 6 x + 1 = 0 x^3-4x^2-6x+1=0 . Find the numerical value of the expression:

p 5 ( q p ) ( p r ) + q 5 ( r q ) ( q p ) + r 5 ( p r ) ( r q ) \frac{p^5}{(q-p)(p-r)} + \frac{q^5}{(r-q)(q-p)} + \frac{r^5}{(p-r)(r-q)}


The answer is -111.

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2 solutions

Kobe Mercado
Apr 21, 2020

Combining those 3 fractional expressions will yield:

p 5 ( r q ) + q 5 ( p r ) + r 5 ( q p ) ( q p ) ( p r ) ( r q ) \frac{p^5(r-q)+q^5(p-r)+r^5(q-p)}{(q-p)(p-r)(r-q)} .

The numerator of the resulting fraction can be factored into ( q p ) ( p r ) ( q r ) ( p ³ + q ³ + r ³ + p ² q + p ² r + q ² p + q ² r + p r ² + q r ² + p q r ) (q-p)(p-r)(q-r)(p³+q³+r³+p²q+p²r+q²p+q²r+pr²+qr²+pqr) .

Upon simplifying, the expression is equivalent to ( p ³ + q ³ + r ³ + p ² q + p ² r + q ² p + q ² r + p r ² + q r ² + p q r ) = [ p ³ + q ³ + r ³ + p ² ( q + r ) + q ² ( p + r ) + r ² ( p + q ) + p q r ] -(p³+q³+r³+p²q+p²r+q²p+q²r+pr²+qr²+pqr)=-[p³+q³+r³+p²(q+r)+q²(p+r)+r²(p+q)+pqr] .

By Vieta's Theorem, p + q + r = 4 q + r = 4 p p+q+r=4 \implies q+r=4-p . Similarly, p + r = 4 q p+r=4-q and p + q = 4 r p+q=4-r .

Therefore, \(-[p³+q³+r³+p²(q+r)+q²(p+r)+r²(p+q)+pqr]=-[p³+q³+r³+p²(4-p)+q²(4-q)+r²(4-r)+pqr]

=-[p³+q³+r³-p³-q³-r³+4(p²+q²+r²)+pqr]

=-4(p²+q²+r²)-pqr\).

By Vieta's Theorem, p + q + r = 4 p+q+r=4 and \(pq+qr+pr=-6 \implies p²+q²+r²=(p+q+r)²-2(pq+qr+pr)

=4²-2(-6)=

16+12=28\). In addition, by Vieta's Theorem, p q r = 1 pqr=-1 .

Thus, the expression is equal to:

4 ( 28 ) ( 1 ) = 112 + 1 = 111 -4(28)-(-1)=-112+1=\boxed{-111} .

@Kobe Mercado , I have found a simpler solution.

Chew-Seong Cheong - 1 year, 1 month ago

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Nice. Thank you! 😁

Kobe Mercado - 1 year, 1 month ago
Chew-Seong Cheong
Apr 21, 2020

Easier solution

From x 3 4 x 2 6 x + 1 = 0 x^3-4x^2-6x + 1 = 0 , we have:

x 3 = 4 x 2 + 6 x 1 x 4 = 4 x 3 + 6 x 2 x = 4 ( 4 x 2 + 6 x 1 ) + 6 x 2 x = 22 x 2 + 23 x 4 x 5 = 22 x 3 + 23 x 2 4 x = 22 ( 4 x 2 + 6 x 1 ) + 23 x 2 4 x = 111 x 2 + 128 x 22 \begin{aligned} x^3 & = 4x^2+6x-1 \\ x^4 & = 4x^3 + 6x^2 - x = 4(4x^2+6x-1) + 6x^2 - x = 22x^2 + 23x - 4 \\ x^5 & = 22x^3 + 23x^2 - 4x = 22(4x^2+6x-1) + 23x^2 - 4x = 111x^2 + 128x - 22 \end{aligned}

Since p p , q q , and r r are the roots, they all satisfy the equations above. Then:

X = p 5 ( p r ) ( q p ) + q 5 ( q p ) ( r q ) + r 5 ( r q ) ( p r ) = p 5 ( r q ) + q 5 ( p r ) + r 5 ( q p ) ( q p ) ( r q ) ( p r ) = 111 ( r p 2 p 2 q + p q 2 q 2 r + q r 2 r 2 p ) + 128 ( r p p q + q r r p + q r r p ) 22 ( r p + p r + q p ) p 2 q + q 2 r + r 2 p p q 2 q r 2 r p 2 = 111 ( p q 2 + q r 2 + r p 2 p 2 q q 2 r r 2 p ) + 128 ( 0 ) 22 ( 0 ) p 2 q + q 2 r + r 2 p p q 2 q r 2 r p 2 = 111 \begin{aligned} X & = \frac {p^5}{(p-r)(q-p)} + \frac {q^5}{(q-p)(r-q)} + \frac {r^5}{(r-q)(p-r)} \\ & = \frac {p^5(r-q) + q^5(p-r) + r^5(q-p)}{(q-p)(r-q)(p-r)} \\ & = \frac {111(rp^2-p^2q+pq^2-q^2r+qr^2-r^2p)+128(rp-pq+qr-rp+qr-rp)-22(r-p+p-r+q-p)}{p^2q+q^2r+r^2p-pq^2-qr^2-rp^2} \\ & = \frac {111(pq^2+qr^2+rp^2-p^2q-q^2r-r^2p)+128(0)-22(0)}{p^2q+q^2r+r^2p-pq^2-qr^2-rp^2} \\ & = \boxed{-111} \end{aligned}


Similar solution as @Kobe Mercado's

Note that by Vieta's formula , we have p + q + r = 4 p+q+r = 4 , p q + q r + r p = 6 pq+qr+rp = -6 , and p q r = 1 pqr = -1 .

X = p 5 ( p r ) ( q p ) + q 5 ( q p ) ( r q ) + r 5 ( r q ) ( p r ) = p 5 ( r q ) + q 5 ( p r ) + r 5 ( q p ) ( q p ) ( r q ) ( p r ) = ( q p ) ( r q ) ( p r ) ( p 3 + q 3 + r 3 + p 2 q + p 2 r + q 2 p + q 2 r + p r 2 + q r 2 + p q r ) ( q p ) ( r q ) ( p r ) = ( p 3 + q 3 + r 3 + p 2 q + p 2 r + q 2 p + q 2 r + p r 2 + q r 2 + p q r ) = ( ( p + q + r ) ( ( p + q + r ) 2 3 ( p q + q r + r p ) ) + 3 p q r + p q ( p + q ) + q r ( q + r ) + r p ( r + p ) + p q r ) = ( ( p + q + r ) ( ( p + q + r ) 2 3 ( p q + q r + r p ) ) + ( p q + q r + r p ) ( p + q + r ) + p q r ) = 4 ( 16 + 18 ) ( 6 ) ( 4 ) + 1 = 111 \begin{aligned} X & = \frac {p^5}{(p-r)(q-p)} + \frac {q^5}{(q-p)(r-q)} + \frac {r^5}{(r-q)(p-r)} \\ & = \frac {p^5(r-q) + q^5(p-r) + r^5(q-p)}{(q-p)(r-q)(p-r)} \\ & = \frac {-\cancel{(q-p)(r-q)(p-r)}(p^3+q^3+r^3+p^2q+p^2r+q^2p+q^2r+pr^2+qr^2+pqr)}{\cancel{(q-p)(r-q)(p-r)}} \\ & = -\left(\blue{p^3+q^3+r^3}+p^2q+p^2r+q^2p+q^2r+pr^2+qr^2+pqr\right) \\ & = -\left(\blue{(p+q+r)\left((p+q+r)^2 - 3(pq+qr+rp)\right) + 3pqr} + pq(p+q)+qr(q+r)+rp(r+p) + pqr\right) \\ & = -\left((p+q+r)\left((p+q+r)^2 - 3(pq+qr+rp)\right) + (pq+qr+rp)(p+q+r)+ pqr\right) \\ & = -4(16+18) - (-6)(4) + 1 \\ & = \boxed{-111} \end{aligned}

You forgot the negative sign after you cancelled out the common factors in the third line of your solution because r-q and q-r are additive inverses.

Kobe Mercado - 1 year, 1 month ago

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Thanks. I have changed it.

Chew-Seong Cheong - 1 year, 1 month ago

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