Question

divide. if the polynomial does not divide evenly, include the remainder as a fraction.\\((x^{3}-8x^{2}+12x)\div(x - 2)\\)

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(x^3 - 8x^2 + 12x\) by the leading term of the divisor \(x - 2\). The leading term of the dividend is \(x^3\) and the leading term of the divisor is \(x\), so \(\frac{x^3}{x}=x^2\). Multiply the divisor \(x - 2\) by \(x^2\) to get \(x^3 - 2x^2\). Subtract this from the dividend:
\[

$$\begin{align*} &(x^3 - 8x^2 + 12x)-(x^3 - 2x^2)\\ =&x^3 - 8x^2 + 12x - x^3 + 2x^2\\ =& - 6x^2 + 12x \end{align*}$$

\]

Step2: Divide the new leading term

Now, divide the leading term of \(-6x^2 + 12x\) (which is \(-6x^2\)) by the leading term of the divisor \(x\), so \(\frac{-6x^2}{x}=-6x\). Multiply the divisor \(x - 2\) by \(-6x\) to get \(-6x^2 + 12x\). Subtract this from \(-6x^2 + 12x\):
\[

$$\begin{align*} &(-6x^2 + 12x)-(-6x^2 + 12x)\\ =& - 6x^2 + 12x + 6x^2 - 12x\\ =& 0 \end{align*}$$

\]
Since there is no remainder left, the division is exact.

Answer:

\(x^2 - 6x\)