Dividing Polynomials worksheet with ten problems requiring polynomial division and fractional remainders.

Problem: Dividing Polynomials

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

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#### 1. $(-3y^3 + 16y^2 + 3y - 10) \div (y - 3)$

Step 1: Set up the division.
- Dividend: $-3y^3 + 16y^2 + 3y - 10$
- Divisor: $y - 3$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-3y^3$) by the leading term of the divisor ($y$):
$$
\frac{-3y^3}{y} = -3y^2
$$
Write $-3y^2$ above the division bar.

2. Multiply $-3y^2$ by the divisor $y - 3$:
$$
-3y^2 \cdot (y - 3) = -3y^3 + 9y^2
$$
Subtract this from the original polynomial:
$$
(-3y^3 + 16y^2 + 3y - 10) - (-3y^3 + 9y^2) = 7y^2 + 3y - 10
$$

3. Repeat the process with the new polynomial $7y^2 + 3y - 10$:
Divide the leading term ($7y^2$) by the leading term of the divisor ($y$):
$$
\frac{7y^2}{y} = 7y
$$
Write $7y$ above the division bar.

4. Multiply $7y$ by the divisor $y - 3$:
$$
7y \cdot (y - 3) = 7y^2 - 21y
$$
Subtract this from the current polynomial:
$$
(7y^2 + 3y - 10) - (7y^2 - 21y) = 24y - 10
$$

5. Repeat the process with the new polynomial $24y - 10$:
Divide the leading term ($24y$) by the leading term of the divisor ($y$):
$$
\frac{24y}{y} = 24
$$
Write $24$ above the division bar.

6. Multiply $24$ by the divisor $y - 3$:
$$
24 \cdot (y - 3) = 24y - 72
$$
Subtract this from the current polynomial:
$$
(24y - 10) - (24y - 72) = 62
$$

The quotient is $-3y^2 + 7y + 24$, and the remainder is $62$. Express the remainder as a fraction:
$$
\text{Quotient: } -3y^2 + 7y + 24, \quad \text{Remainder: } \frac{62}{y - 3}
$$

Final Answer for 1:
$$
\boxed{-3y^2 + 7y + 24 + \frac{62}{y - 3}}
$$

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#### 2. $(-a^2 - 17) \div (d + 8)$

Step 1: Notice that the variable in the divisor is $d$, but the dividend involves $a$. This suggests a typo. Assuming the problem meant to use the same variable consistently, let's proceed with $a$ instead of $d$.

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-a^2$) by the leading term of the divisor ($a$):
$$
\frac{-a^2}{a} = -a
$$
Write $-a$ above the division bar.

2. Multiply $-a$ by the divisor $a + 8$:
$$
-a \cdot (a + 8) = -a^2 - 8a
$$
Subtract this from the original polynomial:
$$
(-a^2 - 17) - (-a^2 - 8a) = 8a - 17
$$

3. The degree of the remainder ($8a - 17$) is less than the degree of the divisor ($a + 8$), so we stop here.

The quotient is $-a$, and the remainder is $8a - 17$. Express the remainder as a fraction:
$$
\text{Quotient: } -a, \quad \text{Remainder: } \frac{8a - 17}{a + 8}
$$

Final Answer for 2:
$$
\boxed{-a + \frac{8a - 17}{a + 8}}
$$

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#### 3. $(-4y^2 + 20y - 18) \div (y + 9)$

Step 1: Set up the division.
- Dividend: $-4y^2 + 20y - 18$
- Divisor: $y + 9$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-4y^2$) by the leading term of the divisor ($y$):
$$
\frac{-4y^2}{y} = -4y
$$
Write $-4y$ above the division bar.

2. Multiply $-4y$ by the divisor $y + 9$:
$$
-4y \cdot (y + 9) = -4y^2 - 36y
$$
Subtract this from the original polynomial:
$$
(-4y^2 + 20y - 18) - (-4y^2 - 36y) = 56y - 18
$$

3. Divide the leading term of the new polynomial ($56y$) by the leading term of the divisor ($y$):
$$
\frac{56y}{y} = 56
$$
Write $56$ above the division bar.

4. Multiply $56$ by the divisor $y + 9$:
$$
56 \cdot (y + 9) = 56y + 504
$$
Subtract this from the current polynomial:
$$
(56y - 18) - (56y + 504) = -522
$$

The quotient is $-4y + 56$, and the remainder is $-522$. Express the remainder as a fraction:
$$
\text{Quotient: } -4y + 56, \quad \text{Remainder: } \frac{-522}{y + 9}
$$

Final Answer for 3:
$$
\boxed{-4y + 56 - \frac{522}{y + 9}}
$$

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#### 4. $(-3n^2 - 10n - 17) \div (n + 3)$

Step 1: Set up the division.
- Dividend: $-3n^2 - 10n - 17$
- Divisor: $n + 3$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-3n^2$) by the leading term of the divisor ($n$):
$$
\frac{-3n^2}{n} = -3n
$$
Write $-3n$ above the division bar.

2. Multiply $-3n$ by the divisor $n + 3$:
$$
-3n \cdot (n + 3) = -3n^2 - 9n
$$
Subtract this from the original polynomial:
$$
(-3n^2 - 10n - 17) - (-3n^2 - 9n) = -n - 17
$$

3. Divide the leading term of the new polynomial ($-n$) by the leading term of the divisor ($n$):
$$
\frac{-n}{n} = -1
$$
Write $-1$ above the division bar.

4. Multiply $-1$ by the divisor $n + 3$:
$$
-1 \cdot (n + 3) = -n - 3
$$
Subtract this from the current polynomial:
$$
(-n - 17) - (-n - 3) = -14
$$

The quotient is $-3n - 1$, and the remainder is $-14$. Express the remainder as a fraction:
$$
\text{Quotient: } -3n - 1, \quad \text{Remainder: } \frac{-14}{n + 3}
$$

Final Answer for 4:
$$
\boxed{-3n - 1 - \frac{14}{n + 3}}
$$

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#### 5. $(b^2 - 20b + 5) \div (b - 9)$

Step 1: Set up the division.
- Dividend: $b^2 - 20b + 5$
- Divisor: $b - 9$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($b^2$) by the leading term of the divisor ($b$):
$$
\frac{b^2}{b} = b
$$
Write $b$ above the division bar.

2. Multiply $b$ by the divisor $b - 9$:
$$
b \cdot (b - 9) = b^2 - 9b
$$
Subtract this from the original polynomial:
$$
(b^2 - 20b + 5) - (b^2 - 9b) = -11b + 5
$$

3. Divide the leading term of the new polynomial ($-11b$) by the leading term of the divisor ($b$):
$$
\frac{-11b}{b} = -11
$$
Write $-11$ above the division bar.

4. Multiply $-11$ by the divisor $b - 9$:
$$
-11 \cdot (b - 9) = -11b + 99
$$
Subtract this from the current polynomial:
$$
(-11b + 5) - (-11b + 99) = -94
$$

The quotient is $b - 11$, and the remainder is $-94$. Express the remainder as a fraction:
$$
\text{Quotient: } b - 11, \quad \text{Remainder: } \frac{-94}{b - 9}
$$

Final Answer for 5:
$$
\boxed{b - 11 - \frac{94}{b - 9}}
$$

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#### 6. $(-2d^2 + 17) \div (d + 5)$

Step 1: Set up the division.
- Dividend: $-2d^2 + 17$
- Divisor: $d + 5$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-2d^2$) by the leading term of the divisor ($d$):
$$
\frac{-2d^2}{d} = -2d
$$
Write $-2d$ above the division bar.

2. Multiply $-2d$ by the divisor $d + 5$:
$$
-2d \cdot (d + 5) = -2d^2 - 10d
$$
Subtract this from the original polynomial:
$$
(-2d^2 + 17) - (-2d^2 - 10d) = 10d + 17
$$

3. Divide the leading term of the new polynomial ($10d$) by the leading term of the divisor ($d$):
$$
\frac{10d}{d} = 10
$$
Write $10$ above the division bar.

4. Multiply $10$ by the divisor $d + 5$:
$$
10 \cdot (d + 5) = 10d + 50
$$
Subtract this from the current polynomial:
$$
(10d + 17) - (10d + 50) = -33
$$

The quotient is $-2d + 10$, and the remainder is $-33$. Express the remainder as a fraction:
$$
\text{Quotient: } -2d + 10, \quad \text{Remainder: } \frac{-33}{d + 5}
$$

Final Answer for 6:
$$
\boxed{-2d + 10 - \frac{33}{d + 5}}
$$

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#### 7. $(-r^2 - 12) \div (r + 2)$

Step 1: Set up the division.
- Dividend: $-r^2 - 12$
- Divisor: $r + 2$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-r^2$) by the leading term of the divisor ($r$):
$$
\frac{-r^2}{r} = -r
$$
Write $-r$ above the division bar.

2. Multiply $-r$ by the divisor $r + 2$:
$$
-r \cdot (r + 2) = -r^2 - 2r
$$
Subtract this from the original polynomial:
$$
(-r^2 - 12) - (-r^2 - 2r) = 2r - 12
$$

3. Divide the leading term of the new polynomial ($2r$) by the leading term of the divisor ($r$):
$$
\frac{2r}{r} = 2
$$
Write $2$ above the division bar.

4. Multiply $2$ by the divisor $r + 2$:
$$
2 \cdot (r + 2) = 2r + 4
$$
Subtract this from the current polynomial:
$$
(2r - 12) - (2r + 4) = -16
$$

The quotient is $-r + 2$, and the remainder is $-16$. Express the remainder as a fraction:
$$
\text{Quotient: } -r + 2, \quad \text{Remainder: } \frac{-16}{r + 2}
$$

Final Answer for 7:
$$
\boxed{-r + 2 - \frac{16}{r + 2}}
$$

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#### 8. $(3x^3 - 13x^2 + 7x - 10) \div (x - 3)$

Step 1: Set up the division.
- Dividend: $3x^3 - 13x^2 + 7x - 10$
- Divisor: $x - 3$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($3x^3$) by the leading term of the divisor ($x$):
$$
\frac{3x^3}{x} = 3x^2
$$
Write $3x^2$ above the division bar.

2. Multiply $3x^2$ by the divisor $x - 3$:
$$
3x^2 \cdot (x - 3) = 3x^3 - 9x^2
$$
Subtract this from the original polynomial:
$$
(3x^3 - 13x^2 + 7x - 10) - (3x^3 - 9x^2) = -4x^2 + 7x - 10
$$

3. Divide the leading term of the new polynomial ($-4x^2$) by the leading term of the divisor ($x$):
$$
\frac{-4x^2}{x} = -4x
$$
Write $-4x$ above the division bar.

4. Multiply $-4x$ by the divisor $x - 3$:
$$
-4x \cdot (x - 3) = -4x^2 + 12x
$$
Subtract this from the current polynomial:
$$
(-4x^2 + 7x - 10) - (-4x^2 + 12x) = -5x - 10
$$

5. Divide the leading term of the new polynomial ($-5x$) by the leading term of the divisor ($x$):
$$
\frac{-5x}{x} = -5
$$
Write $-5$ above the division bar.

6. Multiply $-5$ by the divisor $x - 3$:
$$
-5 \cdot (x - 3) = -5x + 15
$$
Subtract this from the current polynomial:
$$
(-5x - 10) - (-5x + 15) = -25
$$

The quotient is $3x^2 - 4x - 5$, and the remainder is $-25$. Express the remainder as a fraction:
$$
\text{Quotient: } 3x^2 - 4x - 5, \quad \text{Remainder: } \frac{-25}{x - 3}
$$

Final Answer for 8:
$$
\boxed{3x^2 - 4x - 5 - \frac{25}{x - 3}}
$$

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#### 9. $(b^2 - 16) \div (b + 3)$

Step 1: Set up the division.
- Dividend: $b^2 - 16$
- Divisor: $b + 3$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($b^2$) by the leading term of the divisor ($b$):
$$
\frac{b^2}{b} = b
$$
Write $b$ above the division bar.

2. Multiply $b$ by the divisor $b + 3$:
$$
b \cdot (b + 3) = b^2 + 3b
$$
Subtract this from the original polynomial:
$$
(b^2 - 16) - (b^2 + 3b) = -3b - 16
$$

3. Divide the leading term of the new polynomial ($-3b$) by the leading term of the divisor ($b$):
$$
\frac{-3b}{b} = -3
$$
Write $-3$ above the division bar.

4. Multiply $-3$ by the divisor $b + 3$:
$$
-3 \cdot (b + 3) = -3b - 9
$$
Subtract this from the current polynomial:
$$
(-3b - 16) - (-3b - 9) = -7
$$

The quotient is $b - 3$, and the remainder is $-7$. Express the remainder as a fraction:
$$
\text{Quotient: } b - 3, \quad \text{Remainder: } \frac{-7}{b + 3}
$$

Final Answer for 9:
$$
\boxed{b - 3 - \frac{7}{b + 3}}
$$

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#### 10. $(-4n^3 + 8n^2 - 19n + 8) \div (n - 7)$

Step 1: Set up the division.
- Dividend: $-4n^3 + 8n^2 - 19n + 8$
- Divisor: $n - 7$

Step 2: Perform polynomial long division.

1. Divide the leading term of the dividend ($-4n^3$) by the leading term of the divisor ($n$):
$$
\frac{-4n^3}{n} = -4n^2
$$
Write $-4n^2$ above the division bar.

2. Multiply $-4n^2$ by the divisor $n - 7$:
$$
-4n^2 \cdot (n - 7) = -4n^3 + 28n^2
$$
Subtract this from the original polynomial:
$$
(-4n^3 + 8n^2 - 19n + 8) - (-4n^3 + 28n^2) = -20n^2 - 19n + 8
$$

3. Divide the leading term of the new polynomial ($-20n^2$) by the leading term of the divisor ($n$):
$$
\frac{-20n^2}{n} = -20n
$$
Write $-20n$ above the division bar.

4. Multiply $-20n$ by the divisor $n - 7$:
$$
-20n \cdot (n - 7) = -20n^2 + 140n
$$
Subtract this from the current polynomial:
$$
(-20n^2 - 19n + 8) - (-20n^2 + 140n) = -159n + 8
$$

5. Divide the leading term of the new polynomial ($-159n$) by the leading term of the divisor ($n$):
$$
\frac{-159n}{n} = -159
$$
Write $-159$ above the division bar.

6. Multiply $-159$ by the divisor $n - 7$:
$$
-159 \cdot (n - 7) = -159n + 1113
$$
Subtract this from the current polynomial:
$$
(-159n + 8) - (-159n + 1113) = -1105
$$

The quotient is $-4n^2 - 20n - 159$, and the remainder is $-1105$. Express the remainder as a fraction:
$$
\text{Quotient: } -4n^2 - 20n - 159, \quad \text{Remainder: } \frac{-1105}{n - 7}
$$

Final Answer for 10:
$$
\boxed{-4n^2 - 20n - 159 - \frac{1105}{n - 7}}
$$

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Final Answers:

1. $\boxed{-3y^2 + 7y + 24 + \frac{62}{y - 3}}$
2. $\boxed{-a + \frac{8a - 17}{a + 8}}$
3. $\boxed{-4y + 56 - \frac{522}{y + 9}}$
4. $\boxed{-3n - 1 - \frac{14}{n + 3}}$
5. $\boxed{b - 11 - \frac{94}{b - 9}}$
6. $\boxed{-2d + 10 - \frac{33}{d + 5}}$
7. $\boxed{-r + 2 - \frac{16}{r + 2}}$
8. $\boxed{3x^2 - 4x - 5 - \frac{25}{x - 3}}$
9. $\boxed{b - 3 - \frac{7}{b + 3}}$
10. $\boxed{-4n^2 - 20n - 159 - \frac{1105}{n - 7}}$