Question

Find the remainder when the polynomial $x^5 x^4 x^3 x^2 x$ is divided by $x^3-4x$.

Answer 1

Answer:

Remainder would be $5x^2+21$

Step-by-step explanation:

Given,

Dividend = $x^5+x^4+x^3+x^2+x$

Divisor = $x^3-4x$

Using long division ( shown below ),

We get,

$\frac{x^5+x^4+x^3+x^2+x}{x^3-4x}=x^2+x+5+\frac{5x^2+21}{x^3-4x}$

Therefore,

Remainder would be $5x^2+21$