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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 $f(x)\ =\ 3x^4\ -\ 9x^3\ +\ x^2\ +\ 15x\ +\ k$. To do: Therefore, Dividend$=x^4+2x^3-2x^2+x-1$ Divisor$=x^2+2x-3$ $x^2+2x-3$)$x^4+2x^3-2x^2+x-1$($x^2+1-15$ $x^4+2x^3-3x^2$ ------------------------------- $x^2+2x-3$ ------------- 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 $x-2$.
The divisor is $3x^2\ -\ 5$.
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 $r(x)$.
$x^2+x-1$
$-x+2$
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