Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
$g(x)\ =\ 2x^2\ –\ x\ +\ 3;\ f(x)\ =\ 6x^5\ −\ x^4\ +\ 4x^3\ –\ 5x^2\ –\ x\ –\ 15$
Given:
$g(x)\ =\ 2x^2\ –\ x\ +\ 3$ and $f(x)\ =\ 6x^5\ −\ x^4\ +\ 4x^3\ –\ 5x^2\ –\ x\ –\ 15$.
To do:
We have to check whether $g(x)$ is a factor of $f(x)$ by applying the division algorithm.
Solution:
On applying the division algorithm,
Dividend$f(x)\ =\ 6x^5\ −\ x^4\ +\ 4x^3\ –\ 5x^2\ –\ x\ –\ 15$
Divisor$g(x)\ =\ 2x^2\ –\ x\ +\ 3$
If $g(x)$ is a factor of $f(x)$ then the remainder on long division should be $0$.
$2x^2-x+3$)$6x^5-x^4+4x^3-5x^2-x-15$($3x^3+x^2-2x-5$
$6x^5-3x^4+9x^3$
-------------------------------------
$2x^4-5x^3-5x^2-x-15$
$2x^4 -x^3 + 3x^2$
--------------------------------------
$-4x^3-8x^2-x-15$
$-4x^3+2x^2-6x$
--------------------------------
$-10x^2+5x-15$
$-10x^2+5x-15$
--------------------------
$0$
Therefore, $g(x)$ is a factor of $f(x)$.
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