Question

1. divide: (6x³ - 11x² - 47x - 20)÷(2x + 1) a. 3x² - 7x - 20 b. 3x² + 7x - 20 c. 3x² - 4x - 20 d. 3x² + 4x - 20

Explanation:

Step1: Use long - division of polynomials

We divide the polynomial \(6x^{3}-11x^{2}-47x - 20\) by \(2x + 1\). First, divide the leading term of the dividend \(6x^{3}\) by the leading term of the divisor \(2x\), we get \(3x^{2}\).
\[

$$\begin{align*} (6x^{3}-11x^{2}-47x - 20)\div(2x + 1)&=3x^{2}+\frac{-11x^{2}-47x - 20-3x^{2}(2x + 1)}{2x + 1}\\ &=3x^{2}+\frac{-11x^{2}-47x - 20-6x^{3}-3x^{2}}{2x + 1}\\ &=3x^{2}+\frac{-14x^{2}-47x - 20}{2x + 1} \end{align*}$$

\]

Step2: Divide the leading term of the new dividend

Divide the leading term \(-14x^{2}\) of the new dividend \(-14x^{2}-47x - 20\) by the leading term \(2x\) of the divisor, we get \(-7x\).
\[

$$\begin{align*} &=3x^{2}-7x+\frac{-47x - 20-(-7x)(2x + 1)}{2x + 1}\\ &=3x^{2}-7x+\frac{-47x - 20 + 14x^{2}+7x}{2x + 1}\\ &=3x^{2}-7x+\frac{14x^{2}-40x - 20}{2x + 1} \end{align*}$$

\]

Step3: Divide the leading term of the new - new dividend

Divide the leading term \(14x^{2}\) of the new - new dividend \(14x^{2}-40x - 20\) by the leading term \(2x\) of the divisor, we get \(7x\).
\[

$$\begin{align*} &=3x^{2}-7x - 20+\frac{-20-(-20)(2x + 1)}{2x + 1}\\ &=3x^{2}-7x - 20+\frac{-20 + 40x+20}{2x + 1}\\ &=3x^{2}-7x - 20 \end{align*}$$

\]

Answer:

A. \(3x^{2}-7x - 20\)