Question

find the quotient for each equation.
(14 points each question)

1. $\frac{-x^{3}-x + 9}{x + 3} = $
2. $\frac{2x + 6x^{2}-28}{3x + 7} = $
3. $\frac{6x^{2}+11x - 35}{-5 + 3x} = $

Explanation:

Step1: Rearrange numerator (Q5)

$$\frac{-x^3 - x + 9}{x+3} = \frac{-x^3 + 0x^2 - x + 9}{x+3}$$

Step2: Polynomial long division (Q5)

Use synthetic division with root $x=-3$:
Coefficients: $-1, 0, -1, 9$

$$\begin{align*} -1 + 3 &= 2\\ 2*(-3) &= -6\\ -6 + (-1) &= -7\\ -7*(-3) &= 21\\ 21 + 9 &= 30 \end{align*}$$

Quotient: $-x^2 + 3x - 7$, remainder $30$
$$\frac{-x^3 - x + 9}{x+3} = -x^2 + 3x - 7 + \frac{30}{x+3}$$

Step3: Rearrange numerator (Q6)

$$\frac{6x^2 + 2x - 28}{3x+7}$$

Step4: Polynomial long division (Q6)

Use synthetic division with root $x=-\frac{7}{3}$:
Coefficients: $6, 2, -28$

$$\begin{align*} 6*(-\frac{7}{3}) &= -14\\ -14 + 2 &= -12\\ -12*(-\frac{7}{3}) &= 28\\ 28 + (-28) &= 0 \end{align*}$$

Quotient: $2x - 4$, remainder $0$
$$\frac{6x^2 + 2x - 28}{3x+7} = 2x - 4$$

Step5: Rearrange numerator/denominator (Q7)

$$\frac{6x^2 + 11x - 35}{3x - 5}$$

Step6: Polynomial long division (Q7)

Use synthetic division with root $x=\frac{5}{3}$:
Coefficients: $6, 11, -35$

$$\begin{align*} 6*(\frac{5}{3}) &= 10\\ 10 + 11 &= 21\\ 21*(\frac{5}{3}) &= 35\\ 35 + (-35) &= 0 \end{align*}$$

Quotient: $2x + 7$, remainder $0$
$$\frac{6x^2 + 11x - 35}{3x - 5} = 2x + 7$$

Answer:

1. $\boldsymbol{-x^2 + 3x - 7 + \frac{30}{x+3}}$
2. $\boldsymbol{2x - 4}$
3. $\boldsymbol{2x + 7}$