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

Revision as of 19:18, 26 September 2022

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

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 $-x^2-3$

@@ Line 11: / Line 11: @@
 For <math>p(0)</math>,
-<math>(0)^{2018} + (0)^{1776}-3(0)^4-3 = ((0)^3 - (0))Q(x) + (A(0)^2 + B(0) + C)</math>
+<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>
@@ Line 17: / Line 17: @@
 For <math>p(1)</math>,
-<math>(1)^{2018} + (1)^{1776}-3(1)^4-3 = ((1)^3 - (1))Q(x) + (A(1)^2 + B(1) + C)</math>
+<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>
@@ Line 25: / Line 25: @@
 For <math>p(-1)</math>
-<math>(-1)^{2018} + (-1)^{1776}-3(-1)^4-3 = ((-1)^3 - (-1))Q(x) + (A(-1)^2 + B(-1) + C)</math>
+<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>

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
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