Solved SYNTHETIC DIVISION WORKSHEET • Dont forget ZERO | Chegg.com - Free Printable

Let’s solve each problem using synthetic division. We’ll check if the binomial divisor is a factor — that means the remainder should be zero.

We’ll go one by one, carefully.

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A. (p⁴ + 5p³ – 11p² – 25p + 29) ÷ (p + 6)

Step 1: Set up synthetic division.
Divisor: p + 6 → root = -6
Coefficients of dividend: 1 (p⁴), 5 (p³), -11 (p²), -25 (p), 29 (constant)

Write them down:
```
-6 | 1 5 -11 -25 29
|
-------------------------------
```

Step 2: Bring down the first coefficient (1).

Multiply 1 × (-6) = -6 → add to next term: 5 + (-6) = -1
Multiply -1 × (-6) = 6 → add to next: -11 + 6 = -5
Multiply -5 × (-6) = 30 → add to next: -25 + 30 = 5
Multiply 5 × (-6) = -30 → add to last: 29 + (-30) = -1

Remainder = -1 → NOT zero → NOT a factor

Answer for A: No

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B. (y⁴ – 8y³ + 10y² + 2y + 4) ÷ (y – 2)

Root = 2
Coefficients: 1, -8, 10, 2, 4

```
2 | 1 -8 10 2 4
| 2 -12 -4 -4
---------------------------
1 -6 -2 -2 0
```

Remainder = 0 → YES, it IS a factor.

Answer for B: Yes

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C. (8v⁵ + 32v⁴ + 5v + 20) ÷ (v + 4)

Wait — missing degrees! v⁵, v, then jumps to v¹ and constant. So we need zeros for v³ and v².

Coefficients: 8 (v⁵), 32 (v⁴), 0 (v³), 0 (v²), 5 (v), 20 (const)
Root = -4

```
-4 | 8 32 0 0 5 20
| -32 0 0 0 -20
----------------------------------------
8 0 0 0 5 0
```

Remainder = 0 → YES, it IS a factor.

Answer for C: Yes

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D. (3x³ – 4x² – 17x + 6) ÷ (3x – 1)

Important: The leading coefficient of divisor is not 1. Rule says: “If zero value is a fraction, divide all coefficients by denominator.”

First, find root: 3x – 1 = 0 → x = 1/3

But since divisor has leading coefficient 3, we must adjust.

Alternative method: Use synthetic division with root 1/3, but then divide final quotient coefficients by 3? Actually, better to use polynomial long division or adjust.

Wait — let’s follow the worksheet hint: “If zero value is a fraction, then divide all coefficients by denominator.” Hmm, maybe they mean: when you get the quotient from synthetic division with fractional root, divide each coefficient by the denominator of the root? Not quite standard.

Actually, correct approach for non-monic divisors in synthetic division:

You can still do synthetic division with root 1/3, but the resulting quotient will have coefficients that need to be divided by the leading coefficient of the divisor (which is 3) to get the actual quotient.

Let’s try:

Root = 1/3
Coefficients: 3, -4, -17, 6

```
1/3 | 3 -4 -17 6
| 1 -1 -6
--------------------------
3 -3 -18 0
```

Remainder = 0 → good.

Now, the quotient from synthetic division is 3x² - 3x - 18, but because we used root 1/3 (from divisor 3x–1), we must divide this entire quotient by 3 to get the real quotient.

So: (3x² - 3x - 18)/3 = x² - x - 6

And remainder is 0 → so yes, (3x–1) is a factor.

Answer for D: Yes

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E. (4v³ + 6v² – 8v – 12) ÷ (2v – 3)

Root: 2v – 3 = 0 → v = 3/2

Again, leading coefficient of divisor is 2 → after synthetic division, we’ll divide quotient coefficients by 2.

Coefficients: 4, 6, -8, -12
Root = 3/2

```
3/2 | 4 6 -8 -12
| 6 18 15
----------------------------
4 12 10 3
```

Remainder = 3 → NOT zero → NOT a factor.

Even if we adjusted quotient, remainder isn’t zero → doesn’t matter.

Answer for E: No

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F. (6z³ + 5z² – 3z – 2) ÷ (2z + 1)

Root: 2z + 1 = 0 → z = -1/2

Leading coefficient of divisor = 2 → again, after synthetic division, divide quotient by 2.

Coefficients: 6, 5, -3, -2
Root = -1/2

```
-1/2 | 6 5 -3 -2
| -3 -1 2
---------------------------
6 2 -4 0
```

Remainder = 0 → good.

Quotient from synthetic: 6z² + 2z - 4 → divide by 2 → 3z² + z - 2

So yes, (2z+1) is a factor.

Answer for F: Yes

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Final Answers:

A. No
B. Yes
C. Yes
D. Yes
E. No
F. Yes