Question

1. \\(3x^3 - x^2 + x - 2\\) divided by \\(3x^2 + x + 1\\)
2. \\(-3x^4 - 2x - 1\\) divided by \\(x - 1\\)

Explanation:

Step1: Use polynomial long division

We divide \(-3x^4 - 2x - 1\) by \(x - 1\). First, note that the dividend can be written as \(-3x^4 + 0x^3 + 0x^2 - 2x - 1\) to include all powers of \(x\).

Divide the leading term of the dividend \(-3x^4\) by the leading term of the divisor \(x\), we get \(-3x^3\). Multiply the divisor \(x - 1\) by \(-3x^3\): \(-3x^3(x - 1)=-3x^4 + 3x^3\).

Subtract this from the dividend: \((-3x^4 + 0x^3 + 0x^2 - 2x - 1)-(-3x^4 + 3x^3)= - 3x^3+0x^2 - 2x - 1\).

Step2: Divide the new leading term

Divide the leading term \(-3x^3\) by \(x\) to get \(-3x^2\). Multiply the divisor by \(-3x^2\): \(-3x^2(x - 1)=-3x^3 + 3x^2\).

Subtract: \((-3x^3+0x^2 - 2x - 1)-(-3x^3 + 3x^2)= - 3x^2 - 2x - 1\).

Step3: Divide the new leading term

Divide \(-3x^2\) by \(x\) to get \(-3x\). Multiply the divisor by \(-3x\): \(-3x(x - 1)=-3x^2 + 3x\).

Subtract: \((-3x^2 - 2x - 1)-(-3x^2 + 3x)= - 5x - 1\).

Step4: Divide the new leading term

Divide \(-5x\) by \(x\) to get \(-5\). Multiply the divisor by \(-5\): \(-5(x - 1)=-5x + 5\).

Subtract: \((-5x - 1)-(-5x + 5)= - 6\).

So the quotient is \(-3x^3 - 3x^2 - 3x - 5\) and the remainder is \(-6\). We can write the division as:

\(\frac{-3x^4 - 2x - 1}{x - 1}=-3x^3 - 3x^2 - 3x - 5+\frac{-6}{x - 1}\)

Answer:

The quotient is \(-3x^3 - 3x^2 - 3x - 5\) and the remainder is \(-6\), or \(\frac{-3x^4 - 2x - 1}{x - 1}=-3x^3 - 3x^2 - 3x - 5-\frac{6}{x - 1}\)