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What must be subtracted from the polynomial $f(x)\ =\ x^4\ +\ 2x^3\ -\ 13x^2\ -12x\ +\ 21$ so that the resulting polynomial is exactly divisible by $x^2\ -\ 4x\ +\ 3$?
Given:
Given polynomial is $f(x)\ =\ x^4\ +\ 2x^3\ -\ 13x^2\ -12x\ +\ 21$. The divisor is $x^2\ -\ 4x\ +\ 3$.
To do:
We have to find the polynomial that must be added to the polynomial $f(x)\ =\ x^4\ +\ 2x^3\ -\ 13x^2\ -12x\ +\ 21$ so that the resulting polynomial is exactly divisible by $x^2\ -\ 4x\ +\ 3$.
Solution:
Let the remainder when $x^2\ -\ 4x\ +\ 3$ divides $f(x)\ =\ x^4\ +\ 2x^3\ -\ 13x^2\ -12x\ +\ 21$ be $r(x)$.
Therefore,
Dividend$=x^4+2x^3-13x^2-12x+21$
Divisor$=x^2-4x+3$
$x^2-4x+3$)$x^4+2x^3-13x^2-12x+21$($x^2+6x+8$
$x^4-4x^3+3x^2$
------------------------------------
$6x^3-16x^2-12x+21$
$6x^3-24x^2+18x$
--------------------------
$8x^2-30x+21$
$8x^2-32x+24$
-----------------------
$2x-3$
Remainder$r(x)=2x-3$
If we subtract the remainder from the dividend then it is completely divisible by the divisor.
The polynomial that must be added to the polynomial $f(x)\ =\ x^4\ +\ 2x^3\ -\ 13x^2\ -12x\ +\ 21$ so that the resulting polynomial is exactly divisible by $x^2\ -\ 4x\ +\ 3$ is $2x-3$.
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