# What must be added to the polynomial $f(x)\ =\ x^4\ +\ 2x^3\ -\ 2x^2\ +\ x\ -\ 1$ so that the resulting polynomial is exactly divisible by $x^2\ +\ 2x\ -\ 3$?

Given:

Given polynomial is $f(x)\ =\ 3x^4\ -\ 9x^3\ +\ x^2\ +\ 15x\ +\ k$.

The divisor is $3x^2\ -\ 5$.

To do:

We have to find the polynomial that must be added to the polynomial  $f(x)\ =\ x^4\ +\ 2x^3\ -\ 2x^2\ +\ x\ -\ 1$  so that the resulting polynomial is exactly divisible by  $x^2\ +\ 2x\ -\ 3$.

Solution:

Let the remainder when $x^2\ +\ 2x\ -\ 3$ divides $f(x)\ =\ x^4\ +\ 2x^3\ -\ 2x^2\ +\ x\ -\ 1$ be $r(x)$.

Therefore,

Dividend$=x^4+2x^3-2x^2+x-1$

Divisor$=x^2+2x-3$

$x^2+2x-3$)$x^4+2x^3-2x^2+x-1$($x^2+1-15$

$x^4+2x^3-3x^2$

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$x^2+x-1$

$x^2+2x-3$

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$-x+2$

Remainder$r(x)=-x+2$

If we subtract the remainder from the dividend then it is completely divisible by the divisor.

Therefore, we must add $-r(x)=-(-x+2)=x-2$.

The polynomial that must be added to the polynomial  $f(x)\ =\ x^4\ +\ 2x^3\ -\ 2x^2\ +\ x\ -\ 1$  so that the resulting polynomial is exactly divisible by  $x^2\ +\ 2x\ -\ 3$ is $x-2$.

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Updated on: 10-Oct-2022

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