No ordinary polynomial problem

Algebra Level 5

The fifth degree monic polynomial f ( x ) f(x) satisfies the following conditions.

f ( 1 ) = 721 f ( 2 ) = 1201 f ( 3 ) = 1441 f ( 4 ) = 1261 f ( 5 ) = 673 \begin{aligned} f(1) & = 721 \\ f(2) & = 1201 \\ f(3) & = 1441 \\ f(4) & = 1261 \\ f(5) & = 673 \end{aligned}

What is f ( 6 ) f(6) ?


The answer is 1.

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Chew-Seong Cheong by Chew-Seong Cheong , 63, Malaysia

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3 solutions

Akshat Sharda
Jan 30, 2016

f ( x ) = x 5 + p ( x ) ⇒ p ( x ) = f ( x ) − x 5 m x p ( x ) D 1 ( x ) D 2 ( x ) D 3 ( x ) D 4 ( x ) 1 720 449 − 420 − 570 − 168 2 1169 29 − 990 − 738 − 168 3 1198 − 961 − 1728 − 906 4 237 − 2689 − 2634 5 − 2452 − 5323 6 − 7775 ∴ f ( 6 ) = p ( 6 ) + 6 5 = − 7775 + 7776 = 1 f(x)=x^5+p(x) \Rightarrow p(x)=f(x)-x^5 \\ \begin{array}{c}m x & p(x) & D_{1}(x) & D_{2}(x) & D_{3}(x) & D_{4}(x) \\ 1 & 720 & 449 & -420 & -570 & -168 \\ 2 & 1169 & 29 & -990 & -738 & \color{#3D99F6}{-168} \\ 3 & 1198 & -961 & -1728 & \color{#3D99F6}{-906} \\ 4 & 237 & -2689 & \color{#3D99F6}{-2634} \\ 5 & -2452 & \color{#3D99F6}{-5323} \\ 6 & \color{#3D99F6}{-7775} \end{array} \\ \therefore f(6)=p(6)+6^5 = -7775+7776=\boxed{1}

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Jesse Nieminen
Jul 8, 2018

Faster than method of differences by hand WolframAlpha .

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Carlos Victor
Nov 8, 2015

Taking f(x)=x^5+ax^4+bx^3+cx^2+dx+e , and solving the system we find : f(x)=x^5 - 7x^4 - 25x^3 + 115x^2+ 384x + 253, so f(6)=1.

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Yo these kind of problems get more easy if we use Method of Differences. @Chew-Seong Cheong , @Carlos victor

A Former Brilliant Member - 5 years, 4 months ago

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Can you show the solution?

Chew-Seong Cheong - 5 years, 4 months ago

Thanks for the solution. You should key in backslash open bracket before the formula and backslash close bracket after the formula as for the formula to turn out in LaTex as f ( x ) = x 5 + a x 4 + b x 3 + c x 2 + d x + e f(x) = x^5+ax^4+bx^3+cx^2+dx+e . Try it.

Chew-Seong Cheong - 5 years, 7 months ago

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