Question

1. \\((3x^4 + 17x^3 - 18x^2 - 17x + 6) \div (x^3 + 6x^2 - 5)\\)

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(3x^4\) by the leading term of the divisor \(x^3\) to get the first term of the quotient: \(\frac{3x^4}{x^3} = 3x\).

Step2: Multiply the divisor by \(3x\)

Multiply the entire divisor \(x^3 + 6x^2 - 5\) by \(3x\): \(3x(x^3 + 6x^2 - 5)=3x^4 + 18x^3 - 15x\).

Step3: Subtract this from the dividend

Subtract \(3x^4 + 18x^3 - 15x\) from \(3x^4 + 17x^3 - 18x^2 - 17x + 6\):
\[

$$\begin{align*} &(3x^4 + 17x^3 - 18x^2 - 17x + 6)-(3x^4 + 18x^3 - 15x)\\ =&3x^4 + 17x^3 - 18x^2 - 17x + 6 - 3x^4 - 18x^3 + 15x\\ =& -x^3 - 18x^2 - 2x + 6 \end{align*}$$

\]

Step4: Divide the new leading term

Divide the leading term of the new dividend \(-x^3\) by the leading term of the divisor \(x^3\) to get the next term of the quotient: \(\frac{-x^3}{x^3}=-1\).

Step5: Multiply the divisor by \(-1\)

Multiply the entire divisor \(x^3 + 6x^2 - 5\) by \(-1\): \(-1(x^3 + 6x^2 - 5)=-x^3 - 6x^2 + 5\).

Step6: Subtract this from the new dividend

Subtract \(-x^3 - 6x^2 + 5\) from \(-x^3 - 18x^2 - 2x + 6\):
\[

$$\begin{align*} &(-x^3 - 18x^2 - 2x + 6)-(-x^3 - 6x^2 + 5)\\ =&-x^3 - 18x^2 - 2x + 6 + x^3 + 6x^2 - 5\\ =& -12x^2 - 2x + 1 \end{align*}$$

\]
Since the degree of the remainder \(-12x^2 - 2x + 1\) is less than the degree of the divisor \(x^3 + 6x^2 - 5\), we stop.

The quotient is \(3x - 1\) and the remainder is \(-12x^2 - 2x + 1\). So we can write:
\[
(3x^4 + 17x^3 - 18x^2 - 17x + 6)\div(x^3 + 6x^2 - 5)=3x - 1+\frac{-12x^2 - 2x + 1}{x^3 + 6x^2 - 5}
\]

Answer:

The result of the division is \(3x - 1+\frac{-12x^2 - 2x + 1}{x^3 + 6x^2 - 5}\) (or in quotient - remainder form: quotient \(= 3x - 1\), remainder \(= -12x^2 - 2x + 1\))