This worksheet from NC State's MA 111 course offers essential practice in factoring expressions commonly found when calculating derivatives.

Let's solve each problem from the worksheet step by step, using factoring techniques commonly used in calculus (especially when simplifying derivatives). We'll focus on identifying common factors, applying the distributive property, and simplifying expressions completely.

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Problem 1:

$$
2(x-1)^2 + 5(x-1)
$$

Step 1: Factor out the common term $(x-1)$:
$$
= (x-1)\left[2(x-1) + 5\right]
$$

Step 2: Simplify inside the brackets:
$$
= (x-1)(2x - 2 + 5) = (x-1)(2x + 3)
$$

✔ Answer: $(x-1)(2x+3)$

---

Problem 2:

$$
9x^2(x^4 + 2) + 3x^3(4x^3)
$$

Step 1: Simplify the second term:
$$
3x^3 \cdot 4x^3 = 12x^6
$$

So expression becomes:
$$
9x^2(x^4 + 2) + 12x^6
$$

Step 2: Expand first term:
$$
9x^2 \cdot x^4 + 9x^2 \cdot 2 = 9x^6 + 18x^2
$$

Now add:
$$
9x^6 + 18x^2 + 12x^6 = 21x^6 + 18x^2
$$

Step 3: Factor out GCF $3x^2$:
$$
= 3x^2(7x^4 + 6)
$$

✔ Answer: $3x^2(7x^4 + 6)$

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Problem 3:

$$
3x^2(4x - 12)^2 + x^3(2)(4x - 12)(4)
$$

Step 1: Simplify constants in second term:
$$
x^3 \cdot 2 \cdot 4 \cdot (4x - 12) = 8x^3(4x - 12)
$$

Note: $4x - 12 = 4(x - 3)$, so let’s factor that out later.

But first, rewrite:
$$
3x^2(4x - 12)^2 + 8x^3(4x - 12)
$$

Step 2: Factor out common terms:
Both terms have $x^2$, $(4x - 12)$, and at least one more factor.

Factor out $x^2(4x - 12)$:
$$
= x^2(4x - 12)\left[3(4x - 12) + 8x\right]
$$

Step 3: Simplify inside:
$$
3(4x - 12) = 12x - 36 \\
12x - 36 + 8x = 20x - 36
$$

So:
$$
= x^2(4x - 12)(20x - 36)
$$

Step 4: Factor further:
- $4x - 12 = 4(x - 3)$
- $20x - 36 = 4(5x - 9)$

So:
$$
= x^2 \cdot 4(x - 3) \cdot 4(5x - 9) = 16x^2(x - 3)(5x - 9)
$$

✔ Answer: $16x^2(x - 3)(5x - 9)$

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Problem 4:

$$
5(x^2 + 4)^4 (2x)(x - 2)^4 + (x^2 + 4)^5 (4)(x - 2)^3
$$

Step 1: Identify common factors:
- Both terms have $(x^2 + 4)^4$, $(x - 2)^3$, and constants.

Factor out:
$$
(x^2 + 4)^4 (x - 2)^3 \left[5(2x)(x - 2) + 4(x^2 + 4)\right]
$$

Step 2: Simplify inside:
First part: $5 \cdot 2x \cdot (x - 2) = 10x(x - 2) = 10x^2 - 20x$

Second part: $4(x^2 + 4) = 4x^2 + 16$

Add:
$$
10x^2 - 20x + 4x^2 + 16 = 14x^2 - 20x + 16
$$

So:
$$
= (x^2 + 4)^4 (x - 2)^3 (14x^2 - 20x + 16)
$$

Step 3: Factor out GCF from quadratic:
$14x^2 - 20x + 16$: GCF is 2 → $2(7x^2 - 10x + 8)$

Check if $7x^2 - 10x + 8$ factors:
Discriminant: $(-10)^2 - 4(7)(8) = 100 - 224 = -124 < 0$, so no real roots.

So stop here.

✔ Answer: $(x^2 + 4)^4 (x - 2)^3 \cdot 2(7x^2 - 10x + 8)$

---

Problem 5:

$$
3(x-1)^2(x+2)^4 + 4(x-1)^3(x+2)^3
$$

Step 1: Factor out common terms:
Common: $(x-1)^2$, $(x+2)^3$

So:
$$
= (x-1)^2(x+2)^3 \left[3(x+2) + 4(x-1)\right]
$$

Step 2: Simplify inside:
$$
3(x+2) = 3x + 6 \\
4(x-1) = 4x - 4 \\
Sum: 3x + 6 + 4x - 4 = 7x + 2
$$

✔ Answer: $(x-1)^2(x+2)^3(7x + 2)$

---

Problem 6:

$$
2(x+1)(x-1)^{-2} - 2(x+1)^2(x-1)^{-3}
$$

Step 1: Rewrite with negative exponents:
$$
= \frac{2(x+1)}{(x-1)^2} - \frac{2(x+1)^2}{(x-1)^3}
$$

Step 2: Common denominator is $(x-1)^3$

Rewrite first term:
$$
\frac{2(x+1)(x-1)}{(x-1)^3} - \frac{2(x+1)^2}{(x-1)^3}
$$

Step 3: Combine:
$$
= \frac{2(x+1)(x-1) - 2(x+1)^2}{(x-1)^3}
$$

Factor numerator:
Factor $2(x+1)$:
$$
= \frac{2(x+1)\left[(x-1) - (x+1)\right]}{(x-1)^3}
$$

Simplify inside:
$$
x - 1 - x - 1 = -2
$$

So:
$$
= \frac{2(x+1)(-2)}{(x-1)^3} = \frac{-4(x+1)}{(x-1)^3}
$$

✔ Answer: $\boxed{\dfrac{-4(x+1)}{(x-1)^3}}$

---

Problem 7:

$$
6(x^2 + 3x + 1)(2x + 3)^2 + 6(x^2 + 3x + 1)^2
$$

Step 1: Factor out $6(x^2 + 3x + 1)$:
$$
= 6(x^2 + 3x + 1)\left[(2x + 3)^2 + (x^2 + 3x + 1)\right]
$$

Step 2: Expand $(2x + 3)^2 = 4x^2 + 12x + 9$

Add to $(x^2 + 3x + 1)$:
$$
4x^2 + 12x + 9 + x^2 + 3x + 1 = 5x^2 + 15x + 10
$$

Factor:
$$
= 5(x^2 + 3x + 2) = 5(x+1)(x+2)
$$

So:
$$
= 6(x^2 + 3x + 1) \cdot 5(x+1)(x+2) = 30(x^2 + 3x + 1)(x+1)(x+2)
$$

✔ Answer: $30(x^2 + 3x + 1)(x+1)(x+2)$

---

Problem 8:

$$
60x^3(1 - 3x^2)^5(5x^4 - 1)^2 - 30x(5x^4 - 1)^3(1 - 3x^2)^4
$$

Step 1: Factor out common terms:
- $30x$
- $(1 - 3x^2)^4$
- $(5x^4 - 1)^2$

So:
$$
= 30x(1 - 3x^2)^4(5x^4 - 1)^2 \left[2x^2(1 - 3x^2) - (5x^4 - 1)\right]
$$

Why? Because:
- First term has $60x^3 / 30x = 2x^2$, and $(1 - 3x^2)^5 / (1 - 3x^2)^4 = (1 - 3x^2)$
- Second term has $-1$ times $(5x^4 - 1)^3 / (5x^4 - 1)^2 = -(5x^4 - 1)$

So inside bracket:
$$
2x^2(1 - 3x^2) - (5x^4 - 1) = 2x^2 - 6x^4 - 5x^4 + 1 = -11x^4 + 2x^2 + 1
$$

So:
$$
= 30x(1 - 3x^2)^4(5x^4 - 1)^2(-11x^4 + 2x^2 + 1)
$$

We can write as:
$$
= -30x(1 - 3x^2)^4(5x^4 - 1)^2(11x^4 - 2x^2 - 1)
$$

✔ Answer: $-30x(1 - 3x^2)^4(5x^4 - 1)^2(11x^4 - 2x^2 - 1)$

---

Problem 9:

$$
\frac{2x(x+6)^4 - x^2(4)(x+6)^3}{(x+6)^8}
$$

Step 1: Factor numerator:
Numerator: $2x(x+6)^4 - 4x^2(x+6)^3$

Factor out $2x(x+6)^3$:
$$
= 2x(x+6)^3\left[(x+6) - 2x\right] = 2x(x+6)^3(x + 6 - 2x) = 2x(x+6)^3(-x + 6)
$$

So numerator: $2x(x+6)^3(6 - x)$

Denominator: $(x+6)^8$

So:
$$
= \frac{2x(x+6)^3(6 - x)}{(x+6)^8} = \frac{2x(6 - x)}{(x+6)^5}
$$

✔ Answer: $\boxed{\dfrac{2x(6 - x)}{(x+6)^5}}$

---

Problem 10:

$$
\frac{(x-1)^3 - 3(x-5)(x-1)^2}{(x-1)^6}
$$

Step 1: Factor numerator:
Factor $(x-1)^2$:
$$
= (x-1)^2\left[(x-1) - 3(x-5)\right]
$$

Simplify inside:
$$
x - 1 - 3x + 15 = -2x + 14 = -2(x - 7)
$$

So numerator: $(x-1)^2(-2)(x - 7)$

Denominator: $(x-1)^6$

So:
$$
= \frac{-2(x-1)^2(x - 7)}{(x-1)^6} = \frac{-2(x - 7)}{(x-1)^4}
$$

✔ Answer: $\boxed{\dfrac{-2(x - 7)}{(x-1)^4}}$

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Problem 11:

$$
\frac{(x-1)^2(1 - 2x) - 2(2 + x - x^2)(x - 1)}{(x-1)^4}
$$

Step 1: Note: $2 + x - x^2 = -x^2 + x + 2 = -(x^2 - x - 2) = -(x - 2)(x + 1)$

But better to factor out $(x - 1)$ from numerator.

Numerator:
$$
(x-1)^2(1 - 2x) - 2( -x^2 + x + 2 )(x - 1)
$$

Factor $(x - 1)$:
$$
= (x - 1)\left[(x - 1)(1 - 2x) - 2(-x^2 + x + 2)\right]
$$

Compute inside:
First: $(x - 1)(1 - 2x) = x(1 - 2x) -1(1 - 2x) = x - 2x^2 -1 + 2x = -2x^2 + 3x -1$

Second: $-2(-x^2 + x + 2) = 2x^2 - 2x - 4$

Add:
$$
(-2x^2 + 3x - 1) + (2x^2 - 2x - 4) = x - 5
$$

So numerator: $(x - 1)(x - 5)$

Denominator: $(x - 1)^4$

So:
$$
= \frac{(x - 1)(x - 5)}{(x - 1)^4} = \frac{x - 5}{(x - 1)^3}
$$

✔ Answer: $\boxed{\dfrac{x - 5}{(x - 1)^3}}$

---

Problem 12:

$$
\frac{8(1 - 3x)^2(2x - 1)^3 + 6(2x - 1)^4(1 - 3x)}{(1 - 3x)^4}
$$

Step 1: Factor numerator:
Common: $(1 - 3x)$, $(2x - 1)^3$

Factor:
$$
= (1 - 3x)(2x - 1)^3 \left[8(1 - 3x) + 6(2x - 1)\right]
$$

Simplify inside:
$$
8(1 - 3x) = 8 - 24x \\
6(2x - 1) = 12x - 6 \\
Sum: 8 - 24x + 12x - 6 = -12x + 2 = -2(6x - 1)
$$

So numerator:
$$
= (1 - 3x)(2x - 1)^3 \cdot (-2)(6x - 1)
$$

Denominator: $(1 - 3x)^4$

So:
$$
= \frac{-2(1 - 3x)(2x - 1)^3(6x - 1)}{(1 - 3x)^4} = \frac{-2(2x - 1)^3(6x - 1)}{(1 - 3x)^3}
$$

Note: $1 - 3x = -(3x - 1)$, but leave as is.

✔ Answer: $\boxed{\dfrac{-2(2x - 1)^3(6x - 1)}{(1 - 3x)^3}}$

---

Problem 13:

$$
\frac{3(x+2)^2(x-3)^2 - (x+2)^3(2)(x-3)}{(x-3)^4}
$$

Step 1: Factor numerator:
Common: $(x+2)^2(x-3)$

So:
$$
= (x+2)^2(x-3)\left[3(x-3) - 2(x+2)\right]
$$

Simplify inside:
$$
3x - 9 - 2x - 4 = x - 13
$$

So numerator: $(x+2)^2(x-3)(x - 13)$

Denominator: $(x-3)^4$

So:
$$
= \frac{(x+2)^2(x-3)(x - 13)}{(x-3)^4} = \frac{(x+2)^2(x - 13)}{(x-3)^3}
$$

✔ Answer: $\boxed{\dfrac{(x+2)^2(x - 13)}{(x-3)^3}}$

---

Problem 14:

$$
\frac{3(x+2)^2 - 6x(x+2)}{x+2}
$$

Step 1: Factor numerator:
Factor $(x+2)$:
$$
= (x+2)\left[3(x+2) - 6x\right] = (x+2)(3x + 6 - 6x) = (x+2)(-3x + 6)
$$

So:
$$
= \frac{(x+2)(-3x + 6)}{x+2} = -3x + 6 = -3(x - 2)
$$

✔ Answer: $\boxed{-3(x - 2)}$

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $(x-1)(2x+3)$ |
| 2 | $3x^2(7x^4 + 6)$ |
| 3 | $16x^2(x - 3)(5x - 9)$ |
| 4 | $(x^2 + 4)^4 (x - 2)^3 \cdot 2(7x^2 - 10x + 8)$ |
| 5 | $(x-1)^2(x+2)^3(7x + 2)$ |
| 6 | $\dfrac{-4(x+1)}{(x-1)^3}$ |
| 7 | $30(x^2 + 3x + 1)(x+1)(x+2)$ |
| 8 | $-30x(1 - 3x^2)^4(5x^4 - 1)^2(11x^4 - 2x^2 - 1)$ |
| 9 | $\dfrac{2x(6 - x)}{(x+6)^5}$ |
| 10 | $\dfrac{-2(x - 7)}{(x-1)^4}$ |
| 11 | $\dfrac{x - 5}{(x - 1)^3}$ |
| 12 | $\dfrac{-2(2x - 1)^3(6x - 1)}{(1 - 3x)^3}$ |
| 13 | $\dfrac{(x+2)^2(x - 13)}{(x-3)^3}$ |
| 14 | $-3(x - 2)$ |

Let me know if you'd like any explanation expanded!