Polynomial division problems requiring division by binomials with fractional remainders.

Problem: Divide Polynomials by Binomials

We are tasked with dividing each polynomial by a binomial and expressing any remainders in fractional form. Let's solve each problem step by step.

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#### 1. Divide \( (-2h^3 - 18h^2 - 14h + 8) \div (h + 1) \)

Step 1: Set up the division.
We divide \( -2h^3 - 18h^2 - 14h + 8 \) by \( h + 1 \).

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{-2h^3}{h} = -2h^2
\]
Write \( -2h^2 \) as the first term of the quotient.

2. Multiply \( -2h^2 \) by \( h + 1 \):
\[
-2h^2 \cdot (h + 1) = -2h^3 - 2h^2
\]
Subtract this from the original polynomial:
\[
(-2h^3 - 18h^2 - 14h + 8) - (-2h^3 - 2h^2) = -16h^2 - 14h + 8
\]

3. Repeat the process with the new polynomial \( -16h^2 - 14h + 8 \):
\[
\frac{-16h^2}{h} = -16h
\]
Write \( -16h \) as the next term of the quotient.

4. Multiply \( -16h \) by \( h + 1 \):
\[
-16h \cdot (h + 1) = -16h^2 - 16h
\]
Subtract this from the current polynomial:
\[
(-16h^2 - 14h + 8) - (-16h^2 - 16h) = 2h + 8
\]

5. Repeat the process with the new polynomial \( 2h + 8 \):
\[
\frac{2h}{h} = 2
\]
Write \( 2 \) as the next term of the quotient.

6. Multiply \( 2 \) by \( h + 1 \):
\[
2 \cdot (h + 1) = 2h + 2
\]
Subtract this from the current polynomial:
\[
(2h + 8) - (2h + 2) = 6
\]

The remainder is \( 6 \).

Step 3: Write the final answer.
The quotient is \( -2h^2 - 16h + 2 \) and the remainder is \( 6 \). Express the remainder as a fraction:
\[
\frac{-2h^3 - 18h^2 - 14h + 8}{h + 1} = -2h^2 - 16h + 2 + \frac{6}{h + 1}
\]

Final Answer for Part 1:
\[
\boxed{-2h^2 - 16h + 2 + \frac{6}{h + 1}}
\]

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#### 2. Divide \( (3p^3 - 13p^2 + 18p + 13) \div (p - 6) \)

Step 1: Set up the division.
We divide \( 3p^3 - 13p^2 + 18p + 13 \) by \( p - 6 \).

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{3p^3}{p} = 3p^2
\]
Write \( 3p^2 \) as the first term of the quotient.

2. Multiply \( 3p^2 \) by \( p - 6 \):
\[
3p^2 \cdot (p - 6) = 3p^3 - 18p^2
\]
Subtract this from the original polynomial:
\[
(3p^3 - 13p^2 + 18p + 13) - (3p^3 - 18p^2) = 5p^2 + 18p + 13
\]

3. Repeat the process with the new polynomial \( 5p^2 + 18p + 13 \):
\[
\frac{5p^2}{p} = 5p
\]
Write \( 5p \) as the next term of the quotient.

4. Multiply \( 5p \) by \( p - 6 \):
\[
5p \cdot (p - 6) = 5p^2 - 30p
\]
Subtract this from the current polynomial:
\[
(5p^2 + 18p + 13) - (5p^2 - 30p) = 48p + 13
\]

5. Repeat the process with the new polynomial \( 48p + 13 \):
\[
\frac{48p}{p} = 48
\]
Write \( 48 \) as the next term of the quotient.

6. Multiply \( 48 \) by \( p - 6 \):
\[
48 \cdot (p - 6) = 48p - 288
\]
Subtract this from the current polynomial:
\[
(48p + 13) - (48p - 288) = 301
\]

The remainder is \( 301 \).

Step 3: Write the final answer.
The quotient is \( 3p^2 + 5p + 48 \) and the remainder is \( 301 \). Express the remainder as a fraction:
\[
\frac{3p^3 - 13p^2 + 18p + 13}{p - 6} = 3p^2 + 5p + 48 + \frac{301}{p - 6}
\]

Final Answer for Part 2:
\[
\boxed{3p^2 + 5p + 48 + \frac{301}{p - 6}}
\]

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#### 3. Divide \( (-2c^3 - 19c^2 - 19c - 13) \div (c + 6) \)

Step 1: Set up the division.
We divide \( -2c^3 - 19c^2 - 19c - 13 \) by \( c + 6 \).

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend by the leading term of the divisor:
\[
\frac{-2c^3}{c} = -2c^2
\]
Write \( -2c^2 \) as the first term of the quotient.

2. Multiply \( -2c^2 \) by \( c + 6 \):
\[
-2c^2 \cdot (c + 6) = -2c^3 - 12c^2
\]
Subtract this from the original polynomial:
\[
(-2c^3 - 19c^2 - 19c - 13) - (-2c^3 - 12c^2) = -7c^2 - 19c - 13
\]

3. Repeat the process with the new polynomial \( -7c^2 - 19c - 13 \):
\[
\frac{-7c^2}{c} = -7c
\]
Write \( -7c \) as the next term of the quotient.

4. Multiply \( -7c \) by \( c + 6 \):
\[
-7c \cdot (c + 6) = -7c^2 - 42c
\]
Subtract this from the current polynomial:
\[
(-7c^2 - 19c - 13) - (-7c^2 - 42c) = 23c - 13
\]

5. Repeat the process with the new polynomial \( 23c - 13 \):
\[
\frac{23c}{c} = 23
\]
Write \( 23 \) as the next term of the quotient.

6. Multiply \( 23 \) by \( c + 6 \):
\[
23 \cdot (c + 6) = 23c + 138
\]
Subtract this from the current polynomial:
\[
(23c - 13) - (23c + 138) = -151
\]

The remainder is \( -151 \).

Step 3: Write the final answer.
The quotient is \( -2c^2 - 7c + 23 \) and the remainder is \( -151 \). Express the remainder as a fraction:
\[
\frac{-2c^3 - 19c^2 - 19c - 13}{c + 6} = -2c^2 - 7c + 23 + \frac{-151}{c + 6}
\]

Final Answer for Part 3:
\[
\boxed{-2c^2 - 7c + 23 - \frac{151}{c + 6}}
\]

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Final Answers:

1. \(\boxed{-2h^2 - 16h + 2 + \frac{6}{h + 1}}\)
2. \(\boxed{3p^2 + 5p + 48 + \frac{301}{p - 6}}\)
3. \(\boxed{-2c^2 - 7c + 23 - \frac{151}{c + 6}}\)