Polynomial in 2018

Algebra Level 2

Given a function p : R → R p:\mathbb{R}\rightarrow \mathbb{R} and p ( x ) = a x 5 + b x − 1 p\left(x\right) = ax^5 + bx - 1 with a a , b b is constant. \

If p ( x ) p\left(x\right) divided by ( x − 2018 ) (x - 2018) the remainder is − 2020 -2020 , then what is the remainder of p ( x ) p\left(x\right) if divided by ( x + 2018 ) (x + 2018) ?


The answer is 2018.

Erik Catos Lawijaya by Erik Catos Lawijaya , 19, Indonesia

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1 solution

We can use Remainder Theorem in Polynomial. We have:

p ( 2018 ) = a ( 2018 ) 5 + b ( 2018 ) − 1 = − 2020 p(2018) = \ a(2018)^5 + b(2018) - 1 = -2020 p ( 2018 ) = a ( 2018 ) 5 + b ( 2018 ) = − 2019 p(2018) = \ a(2018)^5 + b(2018) = -2019

So, we can find the remainder of p ( x ) p(x) if divided by ( x + 2018 ) (x + 2018) : p ( − 2018 ) = a ( − 2018 ) 5 + b ( − 2018 ) − 1 p(-2018) = \ a(-2018)^5 + b(-2018) - 1 p ( − 2018 ) = − [ a ( 2018 ) 5 + b ( 2018 ) ] − 1 = − ( − 2019 ) − 1 = 2018 p(-2018) = \ -[a(2018)^5 + b(2018)] - 1 = -(-2019) - 1 = \boxed{2018}

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