Question

1. $(x^4 - 3x^3 - 11x^2 + 3x + 10) div (x - 5)$

Explanation:

Step1: Use Polynomial Long Division

Divide the leading term of the dividend \(x^4 - 3x^3 - 11x^2 + 3x + 10\) by the leading term of the divisor \(x - 5\), which is \(x\). So, \(x^4\div x = x^3\).
Multiply the divisor \(x - 5\) by \(x^3\) to get \(x^4 - 5x^3\).
Subtract this from the dividend:
\[

$$\begin{align*} &(x^4 - 3x^3 - 11x^2 + 3x + 10)-(x^4 - 5x^3)\\ =&x^4 - 3x^3 - 11x^2 + 3x + 10 - x^4 + 5x^3\\ =&2x^3 - 11x^2 + 3x + 10 \end{align*}$$

\]

Step2: Divide the new leading term

Now, divide the leading term of \(2x^3 - 11x^2 + 3x + 10\) (which is \(2x^3\)) by \(x\) to get \(2x^2\).
Multiply the divisor \(x - 5\) by \(2x^2\) to get \(2x^3 - 10x^2\).
Subtract this from \(2x^3 - 11x^2 + 3x + 10\):
\[

$$\begin{align*} &(2x^3 - 11x^2 + 3x + 10)-(2x^3 - 10x^2)\\ =&2x^3 - 11x^2 + 3x + 10 - 2x^3 + 10x^2\\ =&-x^2 + 3x + 10 \end{align*}$$

\]

Step3: Divide the new leading term again

Divide the leading term of \(-x^2 + 3x + 10\) (which is \(-x^2\)) by \(x\) to get \(-x\).
Multiply the divisor \(x - 5\) by \(-x\) to get \(-x^2 + 5x\).
Subtract this from \(-x^2 + 3x + 10\):
\[

$$\begin{align*} &(-x^2 + 3x + 10)-(-x^2 + 5x)\\ =&-x^2 + 3x + 10 + x^2 - 5x\\ =&-2x + 10 \end{align*}$$

\]

Step4: Divide the new leading term once more

Divide the leading term of \(-2x + 10\) (which is \(-2x\)) by \(x\) to get \(-2\).
Multiply the divisor \(x - 5\) by \(-2\) to get \(-2x + 10\).
Subtract this from \(-2x + 10\):
\[

$$\begin{align*} &(-2x + 10)-(-2x + 10)\\ =&-2x + 10 + 2x - 10\\ =&0 \end{align*}$$

\]

Step5: Combine the results

The quotient is the sum of the terms we got in each step: \(x^3 + 2x^2 - x - 2\) and the remainder is \(0\).

Answer:

The result of the division \((x^4 - 3x^3 - 11x^2 + 3x + 10)\div(x - 5)\) is \(x^3 + 2x^2 - x - 2\) with a remainder of \(0\), or simply \(x^3 + 2x^2 - x - 2\) (since the remainder is \(0\)).