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

Let’s solve each problem step by step using synthetic division. Remember:

- Write down the coefficients of the polynomial (include 0 for missing degrees).
- Use the root from the binomial divisor (set it equal to zero → e.g., if divisor is \(x + 3\), use -3).
- Bring down the first coefficient, multiply by the root, add to next coefficient, repeat.
- The last number is the remainder. If it’s 0, then the binomial IS a factor.

---

Problem 1A: \((p^4 + 5p^3 - 11p^2 - 25p + 29) ÷ (p + 6)\)

Divisor: \(p + 6 = 0\) → use -6

Coefficients: 1 (p⁴), 5 (p³), -11 (p²), -25 (p), 29 (constant)

Synthetic Division with -6:

```
-6 | 1 5 -11 -25 29
-6 6 30 -30
----------------------------
1 -1 -5 5 -1 ← Remainder = -1 ≠ 0
```

→ Not a factor.

---

Problem 1B: \((y^4 - 8y^3 + 10y^2 + 2y + 4) ÷ (y - 2)\)

Divisor: \(y - 2 = 0\) → use 2

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

Synthetic Division with 2:

```
2 | 1 -8 10 2 4
2 -12 -4 -4
-----------------------
1 -6 -2 -2 0 ← Remainder = 0 → YES, it's a factor!
```

Quotient: \(y^3 - 6y^2 - 2y - 2\)

---

Problem 1C: \((8v^5 + 32v^4 + 5v + 20) ÷ (v + 4)\)

Wait — missing terms! We need all degrees from v⁵ to constant.

So:
\(8v^5 + 32v^4 + 0v^3 + 0v^2 + 5v + 20\)

Divisor: \(v + 4 = 0\) → use -4

Coefficients: 8, 32, 0, 0, 5, 20

Synthetic Division with -4:

```
-4 | 8 32 0 0 5 20
-32 0 0 0 -20
-------------------------------------
8 0 0 0 5 0 ← Remainder = 0 → YES, factor!
```

Quotient: \(8v^4 + 0v^3 + 0v^2 + 0v + 5 = 8v^4 + 5\)

---

Problem 1D: \((3x^3 - 4x^2 - 17x + 6) ÷ (3x - 1)\)

⚠️ Important: Synthetic division only works when the divisor is monic (leading coefficient = 1). Here, divisor is \(3x - 1\), so we must adjust.

Set \(3x - 1 = 0\) → \(x = \frac{1}{3}\)

We’ll do synthetic division with \(\frac{1}{3}\), but remember: since leading coefficient of divisor is 3, the quotient will be off by a factor of 3. Actually, better approach: divide entire polynomial by 3 after? Or just proceed and interpret carefully.

Actually, standard method: use root \(x = \frac{1}{3}\), perform synthetic division, then the quotient you get is correct *except* that you must divide the resulting coefficients by 3? Wait — no.

Better: Let’s do synthetic division with \(x = \frac{1}{3}\), and accept that the result gives us the quotient as if dividing by \((x - \frac{1}{3})\), but our actual divisor is \(3(x - \frac{1}{3}) = 3x - 1\). So the true quotient is the synthetic quotient divided by 3? Hmm — messy.

Alternative: Factor out the 3 from divisor: write as \(3(x - \frac{1}{3})\), so:

\[
\frac{3x^3 - 4x^2 - 17x + 6}{3x - 1} = \frac{3x^3 - 4x^2 - 17x + 6}{3(x - \frac{1}{3})} = \frac{1}{3} \cdot \left( \text{result of dividing by } x - \frac{1}{3} \right)
\]

So let’s do synthetic division with \(x = \frac{1}{3}\):

Coefficients: 3, -4, -17, 6

```
1/3 | 3 -4 -17 6
1 -1 -6
----------------------
3 -3 -18 0 ← Remainder = 0 → YES, it divides evenly!
```

Now, this means:

\[
3x^3 - 4x^2 - 17x + 6 = (x - \frac{1}{3})(3x^2 - 3x - 18)
\]

But we want to divide by \(3x - 1 = 3(x - \frac{1}{3})\), so:

\[
= \frac{(x - \frac{1}{3})(3x^2 - 3x - 18)}{3(x - \frac{1}{3})} = \frac{3x^2 - 3x - 18}{3} = x^2 - x - 6
\]

✔ Final answer: Quotient = \(x^2 - x - 6\), remainder 0 → yes, it’s a factor.

---

Problem 1E: \((4v^3 + 6v^2 - 8v - 12) ÷ (2v - 3)\)

Again, divisor not monic. Set \(2v - 3 = 0\) → \(v = \frac{3}{2}\)

Do synthetic division with \(v = \frac{3}{2}\)

Coefficients: 4, 6, -8, -12

```
3/2 | 4 6 -8 -12
6 18 15
-------------------------
4 12 10 3 ← Remainder = 3 ≠ 0 → NOT a factor
```

Wait — but let’s double-check calculation:

Step-by-step:

Bring down 4.

Multiply 4 × 3/2 = 6 → add to next coefficient: 6 + 6 = 12

Multiply 12 × 3/2 = 18 → add to -8: -8 + 18 = 10

Multiply 10 × 3/2 = 15 → add to -12: -12 + 15 = 3 → remainder 3.

Correct. So not a factor.

But wait — maybe I made a mistake? Let me try factoring differently or check with substitution.

Plug \(v = 3/2\) into original polynomial:

\(4*(27/8) + 6*(9/4) - 8*(3/2) - 12 = (108/8) + (54/4) - 12 - 12 = 13.5 + 13.5 - 12 - 12 = 27 - 24 = 3\) → same as remainder. Correct.

So remainder is 3 → not a factor.

---

Problem 1F: \((6z^3 + 5z^2 - 3z - 2) ÷ (2z + 1)\)

Divisor: \(2z + 1 = 0\) → \(z = -\frac{1}{2}\)

Synthetic division with \(z = -\frac{1}{2}\)

Coefficients: 6, 5, -3, -2

```
-1/2 | 6 5 -3 -2
-3 -1 2
------------------------
6 2 -4 0 ← Remainder = 0 → YES, factor!
```

Now, similar to before: since divisor is \(2(z + \frac{1}{2}) = 2z + 1\), the quotient from synthetic division is for \((z + \frac{1}{2})\), so we need to divide the quotient by 2.

Synthetic gave us: \(6z^2 + 2z - 4\)

So actual quotient = \(\frac{6z^2 + 2z - 4}{2} = 3z^2 + z - 2\)

Check: \((2z + 1)(3z^2 + z - 2) = 6z^3 + 2z^2 - 4z + 3z^2 + z - 2 = 6z^3 + 5z^2 - 3z - 2\) ✔

So yes, it’s a factor. Quotient: \(3z^2 + z - 2\)

---

## Now Problem 2: Completely FACTOR each polynomial given a known factor. Find all zeros.

2A: \(x^3 + 9x^2 + 23x + 15\); known factor: \(x + 5\)

Use synthetic division with root = -5

Coefficients: 1, 9, 23, 15

```
-5 | 1 9 23 15
-5 -20 -15
--------------------
1 4 3 0 ← remainder 0
```

Quotient: \(x^2 + 4x + 3\)

Factor that: \((x + 1)(x + 3)\)

So full factorization: \((x + 5)(x + 1)(x + 3)\)

Zeros: x = -5, -1, -3

---

2B: \(x^3 - x^2 - 14x + 28\); known factor: \(x - 3\)

Root = 3

Coefficients: 1, -1, -14, 28

```
3 | 1 -1 -14 28
3 6 -24
---------------------
1 2 -8 4 ← remainder 4 ≠ 0?? Wait — that can’t be.

Wait — did I copy right? Polynomial: \(x^3 - x^2 - 14x + 28\), factor \(x - 3\)

Plug in x=3: 27 - 9 - 42 + 28 = (27+28) - (9+42) = 55 - 51 = 4 → indeed remainder 4. But problem says “given a known factor” — contradiction?

Wait — maybe typo? Or perhaps I misread.

Looking back at image: “B. \(x^3 - x^2 - 14x + 28; x - 3\)”

But plugging x=3 gives 4, not 0. So either error in problem, or I miscalculated.

Wait — let me recalculate:

\(3^3 = 27\)

\(- (3)^2 = -9\)

\(-14*3 = -42\)

+28

Total: 27 - 9 = 18; 18 - 42 = -24; -24 + 28 = 4 → yes, remainder 4.

But the problem states “given a known factor”, implying it should divide evenly. Maybe it’s \(x - 2\)? Let’s test x=2:

8 - 4 - 28 + 28 = 4 → still 4.

x= -2: -8 -4 +28 +28 = 44

x=4: 64 - 16 -56 +28 = 20

x= -4: -64 -16 +56 +28 = 4

Hmm. Perhaps it’s \(x - \sqrt{something}\)? No.

Wait — maybe the polynomial is different? Looking at image again...

In the image, it says: “B. \(x^3 - x^2 - 14x + 28; x - 3\)”

But mathematically, it doesn't divide. Unless... perhaps it's a trick? Or maybe I need to proceed anyway? But remainder isn't zero.

Wait — let me try factoring by grouping:

\(x^3 - x^2 - 14x + 28 = x^2(x - 1) -14(x - 2)\) — not helpful.

Group as: \((x^3 - x^2) + (-14x + 28) = x^2(x - 1) -14(x - 2)\) — no common factor.

Perhaps the known factor is wrong? Or maybe it's \(x - 2\)? Try synthetic with 2:

```
2 | 1 -1 -14 28
2 2 -24
---------------
1 1 -12 4 → still 4
```

With 4:

```
4 | 1 -1 -14 28
4 12 -8
--------------
1 3 -2 20
```

With -2:

```
-2 | 1 -1 -14 28
-2 6 16
---------------
1 -3 -8 44
```

None give 0. This is strange.

Wait — perhaps the polynomial is \(x^3 - x^2 - 14x + 24\)? Then at x=3: 27-9-42+24=0. That would make sense.

Or maybe +24 instead of +28? In many textbooks, it's often 24.

Given that, and since the problem says "known factor", I suspect a typo, and it should be +24.

Assume it's \(x^3 - x^2 - 14x + 24\), factor \(x - 3\)

Then synthetic:

```
3 | 1 -1 -14 24
3 6 -24
---------------
1 2 -8 0
```

Quotient: \(x^2 + 2x - 8 = (x + 4)(x - 2)\)

So factors: \((x - 3)(x + 4)(x - 2)\), zeros: 3, -4, 2

But since the image says +28, and we got remainder 4, perhaps the problem has an error. However, for the sake of completing, I'll assume it's a typo and proceed with +24, as otherwise it doesn't work.

Alternatively, maybe the factor is not x-3 but something else. But the problem specifies x-3.

Another possibility: perhaps it's \(x^3 - x^2 - 14x + 28\) and factor is \(x - 2\), but we saw remainder 4.

Wait — let me calculate discriminant or something. Or perhaps use rational root theorem.

Possible rational roots: ±1,2,4,7,14,28

Test x=2: 8 - 4 -28 +28 = 4

x= -2: -8 -4 +28 +28 = 44

x=4: 64 -16 -56 +28 = 20

x=7: 343 -49 -98 +28 = 224

x=1: 1-1-14+28=14

x= -1: -1-1+14+28=40

x=14: too big

No integer roots? But cubic must have at least one real root.

Derivative: 3x^2 -2x -14, discriminant 4 + 168 = 172 >0, so two critical points, but values at integers are never 0.

At x=2.5: (15.625) - (6.25) -35 +28 = 15.625 -6.25 =9.375; 9.375 -35 = -25.625; -25.625+28=2.375

x=2.8: 21.952 - 7.84 -39.2 +28 = (21.952+28)=49.952; (7.84+39.2)=47.04; difference 2.912

x=3: as before 4

All positive? At x=0: 28, x=1:14, x=2:4, x=3:4, x=4:20 — minimum around x=2.5 is about 2.375>0, so no real root? But cubic always has at least one real root.

Wait — let me plot mentally: as x-> -infty, f(x)-> -infty, at x=0, f=28, so there must be a root for x<0.

Try x= -3: -27 -9 +42 +28 = 34

x= -4: -64 -16 +56 +28 = 4

x= -5: -125 -25 +70 +28 = -52

Ah! At x= -5: -125 -25 = -150; +70 = -80; +28 = -52

x= -4: -64 -16 = -80; +56 = -24; +28 = 4

So between x= -5 and -4, it changes from -52 to 4, so root there.

But not rational. So likely, the polynomial is mistyped, and it should be +24 instead of +28.

I think for educational purposes, we'll assume it's +24, as it's a common problem.

So proceeding with \(x^3 - x^2 - 14x + 24\), factor \(x - 3\)

As above: quotient \(x^2 + 2x - 8 = (x+4)(x-2)\)

Factors: (x-3)(x+4)(x-2), zeros: 3, -4, 2

---

2C: \(25x^3 + 150x^2 + 131x + 30\); known factor: \(5x + 3\)

First, set \(5x + 3 = 0\) → \(x = -\frac{3}{5}\)

Synthetic division with \(x = -\frac{3}{5}\)

Coefficients: 25, 150, 131, 30

```
-3/5 | 25 150 131 30
-15 -81 -30
--------------------------
25 135 50 0 ← remainder 0 → good
```

Quotient: \(25x^2 + 135x + 50\)

But since divisor was \(5x + 3 = 5(x + \frac{3}{5})\), the actual quotient is this divided by 5? No.

Similar to before: the synthetic division gives us that:

\[
25x^3 + 150x^2 + 131x + 30 = (x + \frac{3}{5})(25x^2 + 135x + 50)
\]

But we want to divide by \(5x + 3 = 5(x + \frac{3}{5})\), so:

\[
= \frac{(x + \frac{3}{5})(25x^2 + 135x + 50)}{5(x + \frac{3}{5})} = \frac{25x^2 + 135x + 50}{5} = 5x^2 + 27x + 10
\]

Now factor \(5x^2 + 27x + 10\)

Find two numbers that multiply to 5*10=50, add to 27: 25 and 2.

So: \(5x^2 + 25x + 2x + 10 = 5x(x + 5) + 2(x + 5) = (5x + 2)(x + 5)\)

So full factorization: \((5x + 3)(5x + 2)(x + 5)\)

Zeros: x = -3/5, -2/5, -5

---

2D: \(6x^3 + 7x^2 - 1\); known factor: \(2x + 1\)

Set \(2x + 1 = 0\) → \(x = -\frac{1}{2}\)

Synthetic division with \(x = -\frac{1}{2}\)

Coefficients: 6, 7, 0, -1 [note: no x term, so coefficient 0 for x]

```
-1/2 | 6 7 0 -1
-3 -2 1
------------------------
6 4 -2 0 ← remainder 0
```

Quotient: \(6x^2 + 4x - 2\)

Since divisor is \(2x + 1 = 2(x + \frac{1}{2})\), actual quotient is \(\frac{6x^2 + 4x - 2}{2} = 3x^2 + 2x - 1\)

Factor \(3x^2 + 2x - 1\): find two numbers multiply to 3*(-1)= -3, add to 2: 3 and -1.

So: \(3x^2 + 3x - x - 1 = 3x(x + 1) -1(x + 1) = (3x - 1)(x + 1)\)

Full factorization: \((2x + 1)(3x - 1)(x + 1)\)

Zeros: x = -1/2, 1/3, -1

---

## Summary of Answers:

1) Perform divisions:

A. Not a factor (remainder -1)

B. Yes, factor; quotient \(y^3 - 6y^2 - 2y - 2\)

C. Yes, factor; quotient \(8v^4 + 5\)

D. Yes, factor; quotient \(x^2 - x - 6\)

E. Not a factor (remainder 3)

F. Yes, factor; quotient \(3z^2 + z - 2\)

2) Factor completely and find zeros:

A. \((x + 5)(x + 1)(x + 3)\); zeros: -5, -1, -3

B. Assuming typo, should be +24: \((x - 3)(x + 4)(x - 2)\); zeros: 3, -4, 2

C. \((5x + 3)(5x + 2)(x + 5)\); zeros: -3/5, -2/5, -5

D. \((2x + 1)(3x - 1)(x + 1)\); zeros: -1/2, 1/3, -1

For B, since the original has +28 and doesn't divide, but problem states "known factor", I'll go with the corrected version.

Final Answer:

1A: Not a factor
1B: Yes, quotient \( y^3 - 6y^2 - 2y - 2 \)
1C: Yes, quotient \( 8v^4 + 5 \)
1D: Yes, quotient \( x^2 - x - 6 \)
1E: Not a factor
1F: Yes, quotient \( 3z^2 + z - 2 \)

2A: Factors: \( (x+5)(x+1)(x+3) \), Zeros: -5, -1, -3
2B: Factors: \( (x-3)(x+4)(x-2) \), Zeros: 3, -4, 2 (assuming polynomial is \( x^3 - x^2 - 14x + 24 \))
2C: Factors: \( (5x+3)(5x+2)(x+5) \), Zeros: -3/5, -2/5, -5
2D: Factors: \( (2x+1)(3x-1)(x+1) \), Zeros: -1/2, 1/3, -1