Question

1. $(k^3 - 9k - 2) div (k - 3)$

Explanation:

Step1: Use polynomial long division

We divide \(k^3 - 9k - 2\) by \(k - 3\). First, divide the leading term of the dividend \(k^3\) by the leading term of the divisor \(k\), we get \(k^2\).
Multiply the divisor \(k - 3\) by \(k^2\): \(k^2(k - 3)=k^3 - 3k^2\)
Subtract this from the dividend: \((k^3 - 9k - 2)-(k^3 - 3k^2)=3k^2 - 9k - 2\)

Step2: Divide the new leading term

Now, divide the leading term of \(3k^2 - 9k - 2\) (which is \(3k^2\)) by the leading term of the divisor \(k\), we get \(3k\).
Multiply the divisor \(k - 3\) by \(3k\): \(3k(k - 3)=3k^2 - 9k\)
Subtract this from \(3k^2 - 9k - 2\): \((3k^2 - 9k - 2)-(3k^2 - 9k)= - 2\)

Step3: Write the result

The quotient is \(k^2 + 3k\) and the remainder is \(-2\). So we can write the division as:
\(\frac{k^3 - 9k - 2}{k - 3}=k^2 + 3k+\frac{-2}{k - 3}=k^2 + 3k-\frac{2}{k - 3}\)

Answer:

\(k^2 + 3k-\frac{2}{k - 3}\)