Question

find the quotient of $(x^{6}-3x^{5}+2x^{4}+3x^{2}-6x)\div(x-2).$

Explanation:

Step1: Set up polynomial long division

Divide $x^6 - 3x^5 + 2x^4 + 0x^3 + 3x^2 - 6x$ by $x-2$ (insert $0x^3$ for missing term).

Step2: Divide leading terms

$\frac{x^6}{x} = x^5$. Multiply $(x-2)$ by $x^5$: $x^6 - 2x^5$. Subtract from dividend:
$$(x^6 - 3x^5) - (x^6 - 2x^5) = -x^5$$
Bring down $2x^4$: $-x^5 + 2x^4$

Step3: Divide new leading terms

$\frac{-x^5}{x} = -x^4$. Multiply $(x-2)$ by $-x^4$: $-x^5 + 2x^4$. Subtract:
$$(-x^5 + 2x^4) - (-x^5 + 2x^4) = 0$$
Bring down $0x^3$: $0x^3$

Step4: Divide leading term of remainder

$\frac{0x^3}{x} = 0x^2$. Multiply $(x-2)$ by $0x^2$: $0$. Subtract, bring down $3x^2$: $3x^2$

Step5: Divide leading terms

$\frac{3x^2}{x} = 3x$. Multiply $(x-2)$ by $3x$: $3x^2 - 6x$. Subtract:
$$(3x^2 - 6x) - (3x^2 - 6x) = 0$$

Answer:

$x^5 - x^4 + 3x$