What is the remainder when x^4-3x^2+7x+3 is divided by x-2?

2 Answers
Jul 11, 2017

"remainder "=21

Explanation:

"one way of dividing is to use the divisor as a factor in "
"the numerator"

"consider the numerator"

color(red)(x^3)(x-2)color(magenta)(+2x^3)-3x^2+7x+3

=color(red)(x^3)(x-2)color(red)(+2x^2)(x-2)color(magenta)(+4x^2)-3x^2+7x+3

=color(red)(x^3)(x-2)color(red)(+2x^2)(x-2)color(red)(+x)(x-2)color(magenta)(+2x)+7x+3

=color(red)(x^3)(x-2)color(red)(+2x^2)(x-2)color(red)(+x)(x-2)color(red)(+9)(x-2)color(magenta)(+18)+3

=color(red)(x^3)(x-2)color(red)(+2x^2)(x-2)color(red)(+x)(x-2)color(red)(+9)(x-2)+21

"quotient "=color(red)(x^3+2x^2+x+9)," remainder "=21

Jul 12, 2017

21

Explanation:

We can use the Remainder theorem , which states;

"For the Polynomial "P(x) " its remainder on division by "

(x-a) " is "P(a)

proof

(x-a)" a factor of "P(x)

=>P(x)=(x-a)Q(x)+R---(1)

where

Q(x)" is the quotient polynomial"

R=" the remainder"

taking (1)

P(a)=cancel((a-a)Q(a))+R

:.P(a)=R

so we have

P(x)=x^4-3x^2+7x+3

it is to be divided by

(x-2)

:. " we require "P(2)

P(2)=2^4-3xx2^2+7xx2+3

P(2)=16-12+14+3

P(2)=21