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Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 8"

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Let <math>p(x) = x^{2018} + x^{1776}-3x^4-3</math>. Find the remainder when you divide <math>p(x)</math> by <math>x^3-x</math>
 
Let <math>p(x) = x^{2018} + x^{1776}-3x^4-3</math>. Find the remainder when you divide <math>p(x)</math> by <math>x^3-x</math>
  
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By the division algorithm, let <math>p(x)\: {\equiv}\: x^{2018} + x^{1776}-3x^4-3 = (x^3 - x)Q(x) + (Ax^2 + Bx + C)</math> for some quotient <math>Q(x)</math>
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By the division algorithm, let <math>p(x) = x^{2018} + x^{1776}-3x^4-3 \:{\equiv}\: (x^3 - x)Q(x) + (Ax^2 + Bx + C)</math> for some quotient <math>Q(x)</math>
  
 
Consider the roots of <math>x^3-x=0 \implies x=0\: or\: x=1 \:or\: x=-1</math>
 
Consider the roots of <math>x^3-x=0 \implies x=0\: or\: x=1 \:or\: x=-1</math>
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For <math>p(0)</math>,  
 
For <math>p(0)</math>,  
  
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<math>(0)^{2018} + (0)^{1776}-3(0)^4-3 = ((0)^3 - (0))Q(x) + (A(0)^2 + B(0) + C)</math>
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<math>(0)^{2018} + (0)^{1776}-3(0)^4-3 = ((0)^3 - (0))Q(0) + (A(0)^2 + B(0) + C)</math>
  
 
<math>C=-3</math>
 
<math>C=-3</math>
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For <math>p(1)</math>,
 
For <math>p(1)</math>,
  
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<math>(1)^{2018} + (1)^{1776}-3(1)^4-3 = ((1)^3 - (1))Q(x) + (A(1)^2 + B(1) + C)</math>
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<math>(1)^{2018} + (1)^{1776}-3(1)^4-3 = ((1)^3 - (1))Q(1) + (A(1)^2 + B(1) + C)</math>
  
 
<math>1 + 1 - 3 -3 = A + B - 3</math>
 
<math>1 + 1 - 3 -3 = A + B - 3</math>
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<math>A+B=-1</math>
 
<math>A+B=-1</math>
  
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For <math>p(-1)</math>
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For <math>p(-1)</math>,
  
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<math>(-1)^{2018} + (-1)^{1776}-3(-1)^4-3 = ((-1)^3 - (-1))Q(x) + (A(-1)^2 + B(-1) + C)</math>
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<math>(-1)^{2018} + (-1)^{1776}-3(-1)^4-3 = ((-1)^3 - (-1))Q(-1) + (A(-1)^2 + B(-1) + C)</math>
  
 
<math>1 + 1 -3 -3 = A - B + -3</math>
 
<math>1 + 1 -3 -3 = A - B + -3</math>

Latest revision as of 19:19, 26 September 2022

Problem

Let $p(x) = x^{2018} + x^{1776}-3x^4-3$. Find the remainder when you divide $p(x)$ by $x^3-x$

Solution

Let $p(x) = x^{2018} + x^{1776}-3x^4-3$. Find the remainder when you divide $p(x)$ by $x^3-x$

By the division algorithm, let $p(x) = x^{2018} + x^{1776}-3x^4-3 \:{\equiv}\: (x^3 - x)Q(x) + (Ax^2 + Bx + C)$ for some quotient $Q(x)$

Consider the roots of $x^3-x=0 \implies x=0\: or\: x=1 \:or\: x=-1$

For $p(0)$,

$(0)^{2018} + (0)^{1776}-3(0)^4-3 = ((0)^3 - (0))Q(0) + (A(0)^2 + B(0) + C)$

$C=-3$

For $p(1)$,

$(1)^{2018} + (1)^{1776}-3(1)^4-3 = ((1)^3 - (1))Q(1) + (A(1)^2 + B(1) + C)$

$1 + 1 - 3 -3 = A + B - 3$

$A+B=-1$

For $p(-1)$,

$(-1)^{2018} + (-1)^{1776}-3(-1)^4-3 = ((-1)^3 - (-1))Q(-1) + (A(-1)^2 + B(-1) + C)$

$1 + 1 -3 -3 = A - B + -3$

$A-B=-1$

$\therefore \begin{cases} A+B=-1 \\ A-B=-1 \end{cases}$

Solving, we get $A=-1$ and $B=0$

$\therefore\:$ the remainder is $-x^2-3$

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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