Question

(3x⁴ - 5x³ + 9x - 6) ÷ (x - 1)

Explanation:

Step1: Use Polynomial Long Division

Divide \(3x^4 - 5x^3 + 9x - 6\) by \(x - 1\). First term: \(\frac{3x^4}{x}=3x^3\). Multiply \(x - 1\) by \(3x^3\): \(3x^4 - 3x^3\). Subtract from dividend: \((3x^4 - 5x^3 + 9x - 6)-(3x^4 - 3x^3)= -2x^3 + 9x - 6\).

Step2: Next Term in Quotient

Divide \(-2x^3\) by \(x\) to get \(-2x^2\). Multiply \(x - 1\) by \(-2x^2\): \(-2x^3 + 2x^2\). Subtract: \((-2x^3 + 9x - 6)-(-2x^3 + 2x^2)= -2x^2 + 9x - 6\).

Step3: Next Term in Quotient

Divide \(-2x^2\) by \(x\) to get \(-2x\). Multiply \(x - 1\) by \(-2x\): \(-2x^2 + 2x\). Subtract: \((-2x^2 + 9x - 6)-(-2x^2 + 2x)= 7x - 6\).

Step4: Next Term in Quotient

Divide \(7x\) by \(x\) to get \(7\). Multiply \(x - 1\) by \(7\): \(7x - 7\). Subtract: \((7x - 6)-(7x - 7)= 1\).

Answer:

The quotient is \(3x^3 - 2x^2 - 2x + 7\) with a remainder of \(1\), so \(\frac{3x^4 - 5x^3 + 9x - 6}{x - 1}=3x^3 - 2x^2 - 2x + 7+\frac{1}{x - 1}\)