North Carolina State University MA 111 Calculus-Related Factoring worksheet with examples and exercises.

Let's solve each problem from the worksheet step by step, using factoring and simplification techniques. The goal is to factor completely and simplify each expression, often involving common factors, distributive property, and algebraic manipulation.

---

Problem 1:

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

Step 1: Factor out the common term $(x-1)$:
$$
= (x-1)\left[2(x-1) + 5\right]
$$
$$
= (x-1)(2x - 2 + 5) = (x-1)(2x + 3)
$$

✔ Answer: $\boxed{(x-1)(2x+3)}$

---

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:
$$
= 9x^2(x^4 + 2) + 12x^6
$$

Step 2: Expand:
$$
= 9x^6 + 18x^2 + 12x^6 = (9x^6 + 12x^6) + 18x^2 = 21x^6 + 18x^2
$$

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

✔ Answer: $\boxed{3x^2(7x^4 + 6)}$

---

Problem 3:

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

Step 1: Simplify constants:
$$
x^3 \cdot 2 \cdot 4 = 8x^3
$$
So:
$$
= 3x^2(4x - 12)^2 + 8x^3(4x - 12)
$$

Step 2: Factor out common terms:
Note that $(4x - 12) = 4(x - 3)$, so we can write:

But let’s keep it as is for now. Common factors: $x^2$, $(4x - 12)$

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

Now compute inside:
$$
3(4x - 12) = 12x - 36
$$
$$
12x - 36 + 8x = 20x - 36
$$

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

Now 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: $\boxed{16x^2(x - 3)(5x - 9)}$

---

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.
- First term: $5 \cdot 2x = 10x$
- Second term: $4$

Common: $(x^2 + 4)^4$, $(x - 2)^3$

Factor them out:
$$
= (x^2 + 4)^4(x - 2)^3 \left[10x(x - 2) + 4(x^2 + 4)\right]
$$

Now simplify inside:
$$
10x(x - 2) = 10x^2 - 20x
$$
$$
4(x^2 + 4) = 4x^2 + 16
$$
Add:
$$
10x^2 - 20x + 4x^2 + 16 = 14x^2 - 20x + 16
$$

Factor this quadratic:
$$
= 2(7x^2 - 10x + 8)
$$

Check discriminant: $(-10)^2 - 4 \cdot 7 \cdot 8 = 100 - 224 = -124 < 0$ → no real roots → irreducible over reals.

So final form:
$$
= (x^2 + 4)^4(x - 2)^3 \cdot 2(7x^2 - 10x + 8)
$$

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

---

Problem 5:

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

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

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

Simplify inside:
$$
3(x+2) = 3x + 6,\quad 4(x-1) = 4x - 4
\Rightarrow 3x + 6 + 4x - 4 = 7x + 2
$$

✔ Answer: $\boxed{(x-1)^2(x+2)^3(7x+2)}$

---

Problem 6:

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

Step 1: Factor out common terms:
- Both have $2$, $(x+1)$, $(x-1)^{-3}$

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

Because:
- First term: $(x-1)^{-2} = (x-1)^{-3} \cdot (x-1)$
- Second term: $(x+1)^2 \cdot (x-1)^{-3}$

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

Simplify:
$$
(x - 1) - (x + 1) = x - 1 - x - 1 = -2
$$

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

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

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

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

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

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

✔ Answer: $\boxed{30(x^2 + 3x + 1)(x+1)(x+2)}$

---

Problem 8:

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

Step 1: Identify common factors:
- $30x$, $(1 - 3x^2)^4$, $(5x^4 - 1)^2$

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

Explanation:
- First term: $60x^3 / 30x = 2x^2$, and $(1 - 3x^2)^5 / (1 - 3x^2)^4 = (1 - 3x^2)$, and $(5x^4 - 1)^2$ remains.
- Second term: $- (5x^4 - 1)^3 / (5x^4 - 1)^2 = -(5x^4 - 1)$

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

Now simplify inside:
$$
2x^2(1 - 3x^2) = 2x^2 - 6x^4
$$
$$
-(5x^4 - 1) = -5x^4 + 1
$$
Add:
$$
2x^2 - 6x^4 - 5x^4 + 1 = -11x^4 + 2x^2 + 1
$$

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

We can leave it like this or factor the quadratic in $x^2$. Let $u = x^2$:
$$
-11u^2 + 2u + 1
$$

Discriminant: $4 + 44 = 48$, not perfect square → irreducible.

✔ Answer: $\boxed{30x(1 - 3x^2)^4(5x^4 - 1)^2(-11x^4 + 2x^2 + 1)}$

---

Problem 9:

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

Step 1: Factor numerator:
- Both terms have $x(x+6)^3$

Factor:
$$
= \frac{x(x+6)^3\left[2(x+6) - 4x\right]}{(x+6)^8}
$$

Simplify inside:
$$
2(x+6) = 2x + 12,\quad -4x
\Rightarrow 2x + 12 - 4x = -2x + 12 = -2(x - 6)
$$

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

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

---

Problem 10:

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

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

$$
= \frac{(x-1)^2\left[(x-1) - 3(x-5)\right]}{(x-1)^6}
= \frac{(x-1) - 3(x-5)}{(x-1)^4}
$$

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

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

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

---

Problem 11:

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

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

But let’s factor numerator:
- Both terms have $(x-1)$

Factor:
$$
= \frac{(x-1)\left[(x-1)(1 - 2x) - 2(2 + x - x^2)\right]}{(x-1)^4}
= \frac{(x-1)(1 - 2x) - 2(2 + x - x^2)}{(x-1)^3}
$$

Now simplify numerator:
First term: $(x-1)(1 - 2x) = x(1 - 2x) -1(1 - 2x) = x - 2x^2 -1 + 2x = -2x^2 + 3x -1$

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

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

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

✔ 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:
$$
= \frac{(1 - 3x)(2x - 1)^3 \left[8(1 - 3x) + 6(2x - 1)\right]}{(1 - 3x)^4}
= \frac{(2x - 1)^3 \left[8(1 - 3x) + 6(2x - 1)\right]}{(1 - 3x)^3}
$$

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

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

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

Alternatively, write:
$$
= \frac{-2(2x - 1)^3(6x - 1)}{(1 - 3x)^3}
$$

✔ 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: $(x+2)^2(x-3)$

Factor:
$$
= \frac{(x+2)^2(x-3)\left[3(x-3) - 2(x+2)\right]}{(x-3)^4}
= \frac{(x+2)^2\left[3(x-3) - 2(x+2)\right]}{(x-3)^3}
$$

Simplify:
$$
3(x-3) = 3x - 9,\quad -2(x+2) = -2x - 4
\Rightarrow 3x - 9 - 2x - 4 = x - 13
$$

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

✔ 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:
- Common: $3(x+2)$

$$
= \frac{3(x+2)\left[(x+2) - 2x\right]}{x+2}
= 3\left[x + 2 - 2x\right] = 3(-x + 2) = -3x + 6
$$

Or: $3(2 - x)$

✔ Answer: $\boxed{3(2 - x)}$ or $\boxed{-3x + 6}$

---

✔ Final Answers Summary:

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

Let me know if you'd like a PDF version or explanation of any specific step!