Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 8"
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== Solution == | == Solution == | ||
+ | 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> | ||
− | <math> | + | 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> |
+ | |||
+ | Consider the roots of <math>x^3-x=0 \implies x=0\: or\: x=1 \:or\: x=-1</math> | ||
+ | |||
+ | 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>C=-3</math> | ||
+ | |||
+ | 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 + 1 - 3 -3 = A + B - 3</math> | ||
+ | |||
+ | <math>A+B=-1</math> | ||
+ | |||
+ | 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 + 1 -3 -3 = A - B + -3</math> | ||
+ | |||
+ | <math>A-B=-1</math> | ||
+ | |||
+ | <math>\therefore \begin{cases} A+B=-1 \\ A-B=-1 \end{cases}</math> | ||
+ | |||
+ | Solving, we get <math>A=-1</math> and <math>B=0</math> | ||
+ | |||
+ | <math>\therefore\:</math> the remainder is <math>-x^2-3</math> | ||
== See also == | == See also == |
Revision as of 19:17, 26 September 2022
Problem
Let . Find the remainder when you divide by
Solution
Let . Find the remainder when you divide by
By the division algorithm, let for some quotient
Consider the roots of
For ,
For ,
For
Solving, we get and
the remainder is
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 |