Practice worksheet for applying the Remainder and Factor Theorems to polynomial division.

Let's solve each problem step by step using the Remainder Theorem and Factor Theorem, as instructed.

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🔷 Recall:

- Remainder Theorem: If a polynomial $ f(x) $ is divided by $ x - c $, then the remainder is $ f(c) $.
- Factor Theorem: $ x - c $ is a factor of $ f(x) $ if and only if $ f(c) = 0 $.

We will evaluate $ f(c) $ for each case and determine:
- The remainder
- Whether $ x - c $ is a factor

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## ✔ Problem 1:
Divide $ x^4 - 2x^3 + x^2 - 14x + 6 $ by $ x - 3 $

We evaluate $ P(3) $:

$$
P(3) = (3)^4 - 2(3)^3 + (3)^2 - 14(3) + 6
$$
$$
= 81 - 2(27) + 9 - 42 + 6
$$
$$
= 81 - 54 + 9 - 42 + 6
$$
$$
= (81 + 9 + 6) - (54 + 42) = 96 - 96 = 0
$$

So:
- $ P(3) = 0 $
- Remainder = 0
- Since remainder is 0, $ x - 3 $ is a factor

✔ Fill in:
- $ P(3) = (3)^4 - 2(3)^3 + (3)^2 - 14(3) + 6 $
- $ P(3) = 0 $
- Since the remainder is 0, then $ x - 3 $ is a factor of $ x^4 - 2x^3 + x^2 - 14x + 6 $

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## ✔ Problem 2:
Divide $ x^4 - 6x^2 - 13x + 13 $ by $ x - 3 $

Evaluate $ P(3) $:

$$
P(3) = (3)^4 - 6(3)^2 - 13(3) + 13
$$
$$
= 81 - 6(9) - 39 + 13
$$
$$
= 81 - 54 - 39 + 13
$$
$$
= (81 + 13) - (54 + 39) = 94 - 93 = 1
$$

So:
- $ P(3) = 1 $
- Remainder = 1
- Since remainder ≠ 0, $ x - 3 $ is not a factor

✔ Fill in:
- $ P(3) = (3)^4 - 6(3)^2 - 13(3) + 13 $
- $ P(3) = 1 $
- Since the remainder is 1, then $ x - 3 $ is not a factor of $ x^4 - 6x^2 - 13x + 13 $

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## ✔ Problem 3:
Divide $ x^3 + 5x^2 + 4x + 16 $ by $ x + 5 $

Note: Divisor is $ x + 5 = x - (-5) $, so we evaluate $ P(-5) $

$$
P(-5) = (-5)^3 + 5(-5)^2 + 4(-5) + 16
$$
$$
= -125 + 5(25) - 20 + 16
$$
$$
= -125 + 125 - 20 + 16
$$
$$
= 0 - 20 + 16 = -4
$$

So:
- $ P(-5) = -4 $
- Remainder = -4
- Since remainder ≠ 0, $ x + 5 $ is not a factor

⚠️ But the question says "then $ x - 3 $ is..." — this is a mistake in the worksheet!

Wait! The divisor is $ x + 5 $, but the conclusion says "then $ x - 3 $ is..." — that’s incorrect. It should be “then $ x + 5 $ is...”

But let’s assume it's a typo and they meant $ x + 5 $. So we’ll write:

✔ Fill in:
- $ P(-5) = (-5)^3 + 5(-5)^2 + 4(-5) + 16 $
- $ P(-5) = -4 $
- Since the remainder is -4, then $ x + 5 $ is not a factor of $ x^3 + 5x^2 + 4x + 16 $

> ⚠️ Note: The blank says "then $ x - 3 $ is..." — this is incorrect. It should say $ x + 5 $. Possibly a typo in the worksheet.

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## ✔ Problem 4:
Divide $ x^3 - x^2 - 14x + 24 $ by $ x + 4 $

Divisor: $ x + 4 = x - (-4) $, so evaluate $ P(-4) $

$$
P(-4) = (-4)^3 - (-4)^2 - 14(-4) + 24
$$
$$
= -64 - 16 + 56 + 24
$$
$$
= (-64 - 16) + (56 + 24) = -80 + 80 = 0
$$

So:
- $ P(-4) = 0 $
- Remainder = 0
- Since remainder is 0, $ x + 4 $ is a factor

But again, the conclusion says: "then $ x - 3 $ is..." — another typo!

It should say: "then $ x + 4 $ is..."

✔ Fill in:
- $ P(-4) = (-4)^3 - (-4)^2 - 14(-4) + 24 $
- $ P(-4) = 0 $
- Since the remainder is 0, then $ x + 4 $ is a factor of $ x^3 - x^2 - 14x + 24 $

> ⚠️ Again, the worksheet incorrectly says "$ x - 3 $" instead of "$ x + 4 $" — likely a formatting error.

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📝 Final Answers (Completed Worksheet):

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#### 1)
$ P(3) = (3)^4 - 2(3)^3 + (3)^2 - 14(3) + 6 $
$ P(3) = 0 $
Since the remainder is 0, then $ x - 3 $ is a factor of $ x^4 - 2x^3 + x^2 - 14x + 6 $

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#### 2)
$ P(3) = (3)^4 - 6(3)^2 - 13(3) + 13 $
$ P(3) = 1 $
Since the remainder is 1, then $ x - 3 $ is not a factor of $ x^4 - 6x^2 - 13x + 13 $

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#### 3)
$ P(-5) = (-5)^3 + 5(-5)^2 + 4(-5) + 16 $
$ P(-5) = -4 $
Since the remainder is -4, then $ x + 5 $ is not a factor of $ x^3 + 5x^2 + 4x + 16 $

> ✘ Worksheet Error: It says "then $ x - 3 $ is..." → should be $ x + 5 $

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#### 4)
$ P(-4) = (-4)^3 - (-4)^2 - 14(-4) + 24 $
$ P(-4) = 0 $
Since the remainder is 0, then $ x + 4 $ is a factor of $ x^3 - x^2 - 14x + 24 $

> ✘ Worksheet Error: It says "then $ x - 3 $ is..." → should be $ x + 4 $

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✔ Summary Table:

| Problem | Divisor | $ f(c) $ | Remainder | Is it a factor? |
|--------|--------|------------|-----------|------------------|
| 1 | $ x - 3 $ | $ f(3) = 0 $ | 0 | Yes |
| 2 | $ x - 3 $ | $ f(3) = 1 $ | 1 | No |
| 3 | $ x + 5 $ | $ f(-5) = -4 $ | -4 | No |
| 4 | $ x + 4 $ | $ f(-4) = 0 $ | 0 | Yes |

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💡 Tips to Remember:

- Use Remainder Theorem: Plug in $ c $ into the polynomial.
- If result is zero, then $ x - c $ is a factor (Factor Theorem).
- Watch out for signs: $ x + 5 = x - (-5) $ → use $ c = -5 $

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✔ Final Answer (For Student to Fill In):

1)
$ P(3) = (3)^4 - 2(3)^3 + (3)^2 - 14(3) + 6 $
$ P(3) = 0 $
Since the remainder is 0, then $ x - 3 $ is a factor of $ x^4 - 2x^3 + x^2 - 14x + 6 $

2)
$ P(3) = (3)^4 - 6(3)^2 - 13(3) + 13 $
$ P(3) = 1 $
Since the remainder is 1, then $ x - 3 $ is not a factor of $ x^4 - 6x^2 - 13x + 13 $

3)
$ P(-5) = (-5)^3 + 5(-5)^2 + 4(-5) + 16 $
$ P(-5) = -4 $
Since the remainder is -4, then $ x + 5 $ is not a factor of $ x^3 + 5x^2 + 4x + 16 $

4)
$ P(-4) = (-4)^3 - (-4)^2 - 14(-4) + 24 $
$ P(-4) = 0 $
Since the remainder is 0, then $ x + 4 $ is a factor of $ x^3 - x^2 - 14x + 24 $

> ⚠️ Please note: The worksheet has errors in problems 3 and 4 — it says "then $ x - 3 $ is..." when it should be $ x + 5 $ and $ x + 4 $ respectively. This might be a printing mistake.

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