Question

When x 4 - 2x 3 - 3x 2 - 4x - 8 is divided by x - 2

Answer 1

Answer:

$\frac{x^4-2x^3-3x^2-4x-8}{x-2}=(x^3-3x-8)-\frac{28}{x-2}$

Step-by-step explanation:

We are given

Dividend is

$=x^4-2x^3-3x^2-4x-8$

Divisor is

$=x-2$

now, we can use long division

so, we get

Quotient as

$Q(x)=x^3-3x-10$

Remainder is

$R=-28$

we can write as

$\frac{x^4-2x^3-3x^2-4x-8}{x-2}=(x^3-3x-8)-\frac{28}{x-2}$