Question

1. $(5x^3 - 7x^2 - 20x - 24) div (x - 3)$
2. $dfrac{6x^4 + 22x^3 - 7x^2 - 29}{x + 4}$

Answer:

Problem 81: \((5x^3 - 7x^2 - 20x - 24) \div (x - 3)\)

We can use polynomial long division or synthetic division to solve this. Let's use synthetic division since the divisor is of the form \(x - c\) (here \(c = 3\)).

Step 1: Set up synthetic division

The coefficients of the dividend \(5x^3 - 7x^2 - 20x - 24\) are \(5, -7, -20, -24\). We use \(c = 3\) for synthetic division.

\[

$$\begin{array}{r|rrrr} 3 & 5 & -7 & -20 & -24 \\ & & 15 & 24 & 12 \\ \hline & 5 & 8 & 4 & -12 \\ \end{array}$$

\]

Wait, that gives a remainder of \(-12\), which means I made a mistake. Let's try polynomial long division.

Step 1: Divide the leading terms

Divide \(5x^3\) by \(x\) to get \(5x^2\). Multiply \((x - 3)\) by \(5x^2\) to get \(5x^3 - 15x^2\). Subtract this from the dividend:

\[

$$\begin{align*} &(5x^3 - 7x^2 - 20x - 24) - (5x^3 - 15x^2) \\ &= 5x^3 - 7x^2 - 20x - 24 - 5x^3 + 15x^2 \\ &= 8x^2 - 20x - 24 \end{align*}$$

\]

Step 2: Divide the new leading term

Divide \(8x^2\) by \(x\) to get \(8x\). Multiply \((x - 3)\) by \(8x\) to get \(8x^2 - 24x\). Subtract this from the new dividend:

\[

$$\begin{align*} &(8x^2 - 20x - 24) - (8x^2 - 24x) \\ &= 8x^2 - 20x - 24 - 8x^2 + 24x \\ &= 4x - 24 \end{align*}$$

\]

Step 3: Divide the new leading term

Divide \(4x\) by \(x\) to get \(4\). Multiply \((x - 3)\) by \(4\) to get \(4x - 12\). Subtract this from the new dividend:

\[

$$\begin{align*} &(4x - 24) - (4x - 12) \\ &= 4x - 24 - 4x + 12 \\ &= -12 \end{align*}$$

\]

Wait, there's a remainder of \(-12\). That suggests maybe the original polynomial is not divisible by \(x - 3\), or I made a mistake in the problem. Wait, maybe the dividend is \(5x^3 - 7x^2 - 20x + 24\)? Let's check with \(x = 3\):

\(5(3)^3 - 7(3)^2 - 20(3) + 24 = 135 - 63 - 60 + 24 = 36\). No. Wait, maybe the dividend is \(5x^3 - 7x^2 - 20x + 24\)? No. Wait, let's check the problem again. The problem is \((5x^3 - 7x^2 - 20x - 24) \div (x - 3)\). Let's try \(x = -1\): \(5(-1)^3 - 7(-1)^2 - 20(-1) - 24 = -5 -7 +20 -24 = -16\). \(x = 4\): \(5(64) -7(16) -20(4) -24 = 320 -112 -80 -24 = 104\). \(x = 2\): \(5(8) -7(4) -20(2) -24 = 40 -28 -40 -24 = -52\). \(x = -2\): \(5(-8) -7(4) -20(-2) -24 = -40 -28 +40 -24 = -52\). Hmm. Maybe the problem has a typo, but assuming we proceed with the division, the quotient is \(5x^2 + 8x + 4\) with a remainder of \(-12\), so \(\frac{5x^3 - 7x^2 - 20x - 24}{x - 3} = 5x^2 + 8x + 4 - \frac{12}{x - 3}\). But maybe I made a mistake in synthetic division. Let's try again:

Synthetic division for \(x = 3\):

Coefficients: \(5, -7, -20, -24\)

Bring down the 5.

Multiply 5 by 3: 15. Add to -7: 8.

Multiply 8 by 3: 24. Add to -20: 4.

Multiply 4 by 3: 12. Add to -24: -12. So the remainder is -12, so the division gives \(5x^2 + 8x + 4\) with remainder -12. So \((5x^3 - 7x^2 - 20x - 24) = (x - 3)(5x^2 + 8x + 4) - 12\).

Problem 82: \(\frac{6x^4 + 22x^3 - 7x^2 - 29}{x + 4}\)

We can use polynomial long division or synthetic division (with \(c = -4\) since the divisor is \(x + 4 = x - (-4)\)).

Step 1: Set up synthetic division

Coefficients of the dividend \(6x^4 + 22x^3 - 7x^2 + 0x - 29\) (we add a 0 for the missing \(x\) term) are \(6, 22, -7, 0, -29\). \(c = -4\).

\[

$$\begin{array}{r|rrrrr} -4 & 6 & 22 & -7 & 0 & -29 \\ & & -24 & 8 & -4 & 16 \\ \hline & 6 & -2 & 1 & -4 & -13 \\ \end{array}$$

\]

So the remainder is -13, and the quotient is \(6x^3 - 2x^2 + x - 4\). Let's verify with polynomial long division.

Step 1: Divide leading terms

Divide \(6x^4\) by \(x\) to get \(6x^3\). Multiply \((x + 4)\) by \(6x^3\) to get \(6x^4 + 24x^3\). Subtract from the dividend:

\[

$$\begin{align*} &(6x^4 + 22x^3 - 7x^2 + 0x - 29) - (6x^4 + 24x^3) \\ &= 6x^4 + 22x^3 - 7x^2 + 0x - 29 - 6x^4 - 24x^3 \\ &= -2x^3 - 7x^2 + 0x - 29 \end{align*}$$

\]

Step 2: Divide new leading term

Divide \(-2x^3\) by \(x\) to get \(-2x^2\). Multiply \((x + 4)\) by \(-2x^2\) to get \(-2x^3 - 8x^2\). Subtract:

\[

$$\begin{align*} &(-2x^3 - 7x^2 + 0x - 29) - (-2x^3 - 8x^2) \\ &= -2x^3 - 7x^2 + 0x - 29 + 2x^3 + 8x^2 \\ &= x^2 + 0x - 29 \end{align*}$$

\]

Step 3: Divide new leading term

Divide \(x^2\) by \(x\) to get \(x\). Multiply \((x + 4)\) by \(x\) to get \(x^2 + 4x\). Subtract:

\[

$$\begin{align*} &(x^2 + 0x - 29) - (x^2 + 4x) \\ &= x^2 + 0x - 29 - x^2 - 4x \\ &= -4x - 29 \end{align*}$$

\]

Step 4: Divide new leading term

Divide \(-4x\) by \(x\) to get \(-4\). Multiply \((x + 4)\) by \(-4\) to get \(-4x - 16\). Subtract:

\[

$$\begin{align*} &(-4x - 29) - (-4x - 16) \\ &= -4x - 29 + 4x + 16 \\ &= -13 \end{align*}$$

\]

So the quotient is \(6x^3 - 2x^2 + x - 4\) with a remainder of \(-13\), so \(\frac{6x^4 + 22x^3 - 7x^2 - 29}{x + 4} = 6x^3 - 2x^2 + x - 4 - \frac{13}{x + 4}\).

Final Answers

1. \(\boldsymbol{5x^2 + 8x + 4 - \frac{12}{x - 3}}\) (or quotient \(5x^2 + 8x + 4\) with remainder \(-12\))

1. \(\boldsymbol{6x^3 - 2x^2 + x - 4 - \frac{13}{x + 4}}\) (or quotient \(6x^3 - 2x^2 + x - 4\) with remainder \(-13\))