Synthetic Division Worksheets - Math Monks - Free Printable

Let’s solve each problem using synthetic division. I’ll go step by step for each one, showing the setup and calculation, then write the final answer.

---

Problem 1: (x³ - 2x² - 5x + 6) ÷ (x - 3)

We already have the synthetic division set up:

Divisor root: 3

Coefficients of dividend: 1 -2 -5 +6

Set up:

```
3 | 1 -2 -5 6
↓ 3 3 -6
---------------------
1 1 -2 0
```

- Bring down the 1.
- Multiply 1 × 3 = 3 → add to -2 → get 1.
- Multiply 1 × 3 = 3 → add to -5 → get -2.
- Multiply -2 × 3 = -6 → add to 6 → get 0.

Remainder is 0 → no remainder.

Quotient coefficients: 1, 1, -2 → which means x² + x - 2

✔ Answer: x² + x - 2

---

Problem 2: (x⁴ - 5x³ + 7x² - 34x - 1) ÷ (x - 5)

Divisor root: 5

Coefficients: 1 -5 7 -34 -1

Synthetic division:

```
5 | 1 -5 7 -34 -1
↓ 5 0 35 5
----------------------------
1 0 7 1 4
```

Steps:

- Bring down 1.
- 1 × 5 = 5 → -5 + 5 = 0
- 0 × 5 = 0 → 7 + 0 = 7
- 7 × 5 = 35 → -34 + 35 = 1
- 1 × 5 = 5 → -1 + 5 = 4 → remainder

So quotient is: x³ + 0x² + 7x + 1 → simplify to x³ + 7x + 1, with remainder 4

But since we’re dividing polynomials, we write as:

x³ + 7x + 1 + 4/(x - 5) — but usually in these problems, if they ask for “Ans” after synthetic division, they want the quotient polynomial and note the remainder separately. However, looking at Problem 1, they just wrote the quotient (since remainder was 0). Here, remainder is not zero.

Wait — let me double-check my math.

Check multiplication/addition:

Start: 1

→ 1×5=5; -5+5=0 ✔️

→ 0×5=0; 7+0=7 ✔️

→ 7×5=35; -34+35=1 ✔️

→ 1×5=5; -1+5=4 ✔️

Yes, remainder is 4.

But in many school worksheets, if there's a remainder, you still write the quotient and mention remainder. But here, the blank says "Ans:" — probably expecting the full result including remainder? Or maybe just the quotient?

Looking back at Problem 1: they gave only the quotient because remainder was 0.

In standard synthetic division answers, if remainder ≠ 0, we write:

Quotient + Remainder / Divisor

But sometimes worksheets expect just the quotient polynomial and assume you know remainder is separate.

However, let’s check the degree: original is degree 4, divisor degree 1 → quotient should be degree 3.

Our quotient is x³ + 0x² + 7x + 1 → so x³ + 7x + 1, remainder 4.

I think for consistency with Problem 1, they might expect just the quotient polynomial, but that would be incomplete. Actually, in most cases, when they say “divide using synthetic division”, and give an “Ans:” line, they expect the full expression including remainder if any.

But let’s see other problems — perhaps we can infer.

Alternatively, maybe I made a mistake? Let me verify by multiplying back.

Multiply (x - 5)(x³ + 7x + 1) + 4

First: (x - 5)(x³ + 7x + 1) = x(x³ + 7x + 1) -5(x³ + 7x + 1)

= x⁴ + 7x² + x -5x³ -35x -5

= x⁴ -5x³ +7x² +x -35x -5 = x⁴ -5x³ +7x² -34x -5

Then add remainder 4: x⁴ -5x³ +7x² -34x -5 +4 = x⁴ -5x³ +7x² -34x -1 → matches original!

So correct.

Now, how to write answer? Since Problem 1 had no remainder and they wrote just the quotient, but here there is a remainder, I think we should write:

x³ + 7x + 1 R 4 or x³ + 7x + 1 + \frac{4}{x-5}

But looking at the format, probably they expect the quotient polynomial and perhaps note remainder. However, in many such worksheets, if remainder exists, they still write the quotient and put remainder aside.

Actually, re-examining the worksheet: in Problem 1, they showed the synthetic division table and then wrote Ans: x² + x - 2 — which is the quotient.

Similarly, for Problem 2, even though remainder is 4, perhaps they still want the quotient polynomial? That doesn’t make sense because it’s not exact.

Wait — maybe I misread the problem? Let me check the original image description.

The user provided text: “(x⁴ - 5x³ + 7x² - 34x - 1) ÷ (x - 5)”

And in synthetic division, we got remainder 4.

But perhaps in the context of this worksheet, they expect the answer as the quotient plus remainder over divisor? But the “Ans:” line is short.

Looking at other problems, like #3: (7x³ +4x +8) ÷ (x+2) — note missing x² term, so coefficient 0.

I think for accuracy, I'll write the quotient and specify remainder.

But to match the style of Problem 1, where they only wrote the quotient (because remainder 0), for Problem 2, since remainder is not 0, I should include it.

Perhaps the expected answer is the entire result: quotient and remainder.

I recall that in some curricula, they write the answer as “quotient with remainder r”.

Given that, I’ll proceed to write for each problem the quotient polynomial and if remainder ≠0, indicate it.

But let’s continue solving all, and see pattern.

---

Problem 3: (7x³ + 4x + 8) ÷ (x + 2)

Note: missing x² term → coefficient 0.

Divisor: x + 2 → root is -2

Coefficients: 7 0 4 8 (since 7x³ + 0x² + 4x + 8)

Synthetic division with -2:

```
-2 | 7 0 4 8
↓ -14 28 -64
---------------------
7 -14 32 -56
```

Steps:

- Bring down 7
- 7 × (-2) = -14 → 0 + (-14) = -14
- -14 × (-2) = 28 → 4 + 28 = 32
- 32 × (-2) = -64 → 8 + (-64) = -56 → remainder

Quotient: 7x² -14x + 32, remainder -56

So answer: 7x² - 14x + 32 with remainder -56

Or written as: 7x² -14x +32 - 56/(x+2)

But again, for "Ans:", probably they want the quotient polynomial? But that ignores remainder.

I think I need to follow the convention: in synthetic division, the last number is remainder, others are coefficients of quotient.

For reporting, if remainder is 0, just quotient; else, quotient plus remainder fraction.

But to be safe, I'll compute all and decide.

---

Problem 4: (x³ + 5x² + 7x + 2) ÷ (x + 2)

Root: -2

Coefficients: 1 5 7 2

Synthetic division:

```
-2 | 1 5 7 2
↓ -2 -6 -2
---------------------
1 3 1 0
```

Steps:

- Bring down 1
- 1×(-2)= -2 → 5 + (-2)=3
- 3×(-2)= -6 → 7 + (-6)=1
- 1×(-2)= -2 → 2 + (-2)=0 → remainder 0

Quotient: x² + 3x + 1

✔ Answer: x² + 3x + 1

---

Problem 5: (3x² + 7x - 20) ÷ (x + 5)

Root: -5

Coefficients: 3 7 -20

Synthetic division:

```
-5 | 3 7 -20
↓ -15 40
---------------
3 -8 20
```

Steps:

- Bring down 3
- 3×(-5)= -15 → 7 + (-15)= -8
- -8×(-5)= 40 → -20 + 40=20 → remainder 20

Quotient: 3x - 8, remainder 20

Answer: 3x - 8 with remainder 20

---

Problem 6: (7x³ + 6x - 8) ÷ (x - 4)

Missing x² term → coefficient 0.

Root: 4

Coefficients: 7 0 6 -8

Synthetic division:

```
4 | 7 0 6 -8
↓ 28 112 472
---------------------
7 28 118 464
```

Steps:

- Bring down 7
- 7×4=28 → 0+28=28
- 28×4=112 → 6+112=118
- 118×4=472 → -8+472=464 → remainder

Quotient: 7x² + 28x + 118, remainder 464

That seems large, but let's verify quickly.

Multiply (x-4)(7x² +28x +118) +464

First: x*(7x²+28x+118) = 7x³ +28x² +118x

-4*(7x²+28x+118) = -28x² -112x -472

Sum: 7x³ + (28x²-28x²) + (118x-112x) -472 = 7x³ +6x -472

Add remainder 464: 7x³ +6x -472 +464 = 7x³ +6x -8 → correct.

So yes.

Answer: 7x² + 28x + 118 with remainder 464

---

Problem 7: (3x⁴ - 5x² + 6) ÷ (x - 2)

Missing x³ and x terms → coefficients: 3 (x⁴), 0 (x³), -5 (x²), 0 (x), 6 (constant)

Root: 2

Coefficients: 3 0 -5 0 6

Synthetic division:

```
2 | 3 0 -5 0 6
↓ 6 12 14 28
---------------------------
3 6 7 14 34
```

Steps:

- Bring down 3
- 3×2=6 → 0+6=6
- 6×2=12 → -5+12=7
- 7×2=14 → 0+14=14
- 14×2=28 → 6+28=34 → remainder

Quotient: 3x³ + 6x² + 7x + 14, remainder 34

Answer: 3x³ + 6x² + 7x + 14 with remainder 34

---

Problem 8: (8x² - 5x + 6) ÷ (x - 2)

Root: 2

Coefficients: 8 -5 6

Synthetic division:

```
2 | 8 -5 6
↓ 16 22
---------------
8 11 28
```

Steps:

- Bring down 8
- 8×2=16 → -5+16=11
- 11×2=22 → 6+22=28 → remainder

Quotient: 8x + 11, remainder 28

Answer: 8x + 11 with remainder 28

---

Now, going back to how to present answers.

In Problem 1, they have Ans: x² + x - 2 (remainder 0)

In Problem 4, similarly, remainder 0, so Ans: x² + 3x + 1

For others, remainder not zero.

I think for consistency, since the worksheet likely expects the quotient polynomial (as the main part), and perhaps the remainder is implied or to be noted, but in the "Ans:" line, they might just want the quotient.

However, that would be misleading for non-zero remainders.

Perhaps in this context, "Ans" means the result of the division, which includes remainder.

But looking at the format, and since Problem 1 shows only quotient, I suspect that for problems with remainder, they still write the quotient and assume student knows to handle remainder.

To be precise, I'll provide the quotient polynomial for each, and if remainder is not zero, I'll mention it in parentheses or something, but the instruction says "provide only the final correct answer" in Final Answer section.

Re-reading user instruction: "Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]"

And for each problem, there is an "Ans:" line.

Also, in the initial example, for Problem 1, they filled in x² + x - 2.

So I think for all problems, they expect the quotient polynomial, and if there is a remainder, it's understood that it's separate, but perhaps not written in the "Ans" box.

But that doesn't make sense for accuracy.

Another idea: in some systems, they write the answer as the quotient, and the remainder is used for verification, but for the answer, it's the quotient.

I recall that in synthetic division, the numbers obtained (except last) are coefficients of quotient, last is remainder.

When they say "divide", the result is quotient plus remainder/divisor.

But for the purpose of this worksheet, since it's titled "Synthetic Division" and they show the process, likely they want the quotient polynomial listed in "Ans".

Moreover, in Problem 2, if I look online or standard practice, sometimes they list the quotient.

Let me check Problem 3: (7x³ +4x +8) ÷ (x+2)

We have quotient 7x² -14x +32, remainder -56.

If I were to write the full answer, it's 7x² -14x +32 - 56/(x+2)

But the "Ans" line is short, so probably they want 7x² -14x +32

Similarly for others.

Perhaps the worksheet assumes that students will write the quotient, and the remainder is for their own check.

To resolve this, I'll assume that "Ans" refers to the quotient polynomial, as in Problem 1.

For problems with remainder, it's still the quotient, and the remainder is additional information.

So for each problem, I'll provide the quotient polynomial as the answer.

Let's list them:

1. x² + x - 2 (remainder 0)

2. x³ + 7x + 1 (remainder 4) — but wait, in our calculation, quotient is x³ + 0x² +7x +1, so x³ +7x +1

3. 7x² -14x +32 (remainder -56)

4. x² +3x +1 (remainder 0)

5. 3x -8 (remainder 20)

6. 7x² +28x +118 (remainder 464)

7. 3x³ +6x² +7x +14 (remainder 34)

8. 8x +11 (remainder 28)

Now, for Problem 2, is x³ +7x +1 correct? Yes, since no x² term.

Similarly for others.

I think this is what is expected.

So final answers per problem:

1. x² + x - 2

2. x³ + 7x + 1

3. 7x² - 14x + 32

4. x² + 3x + 1

5. 3x - 8

6. 7x² + 28x + 118

7. 3x³ + 6x² + 7x + 14

8. 8x + 11

Even though there are remainders, the "Ans" likely wants the quotient polynomial.

To confirm, in Problem 1, they have exactly that.

So I'll go with that.

Final Answer for each:

[1] x² + x - 2
[2] x³ + 7x + 1
[3] 7x² - 14x + 32
[4] x² + 3x + 1
[5] 3x - 8
[6] 7x² + 28x + 118
[7] 3x³ + 6x² + 7x + 14
[8] 8x + 11

Note: For problems with non-zero remainders, the full division result includes the remainder, but based on the worksheet format and Problem 1 example, the "Ans" field expects the quotient polynomial.