Pre-Calculus - Factoring Worksheet featuring problems like 5x² + 8x - 4 and x² - 100 for students to factor.

Let's solve each expression on the Pre-Calculus - Factoring Worksheet step by step. We'll factor each expression completely if possible.

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1. $ 5x^2 + 8x - 4 $

We need to factor this quadratic trinomial.

Use the AC method:

- $ a = 5 $, $ b = 8 $, $ c = -4 $
- $ ac = 5 \cdot (-4) = -20 $
- Find two numbers that multiply to $-20$ and add to $8$:
→ $10$ and $-2$

Now split the middle term:
$$
5x^2 + 10x - 2x - 4
$$

Group:
$$
(5x^2 + 10x) - (2x + 4) = 5x(x + 2) - 2(x + 2)
$$

Factor out common binomial:
$$
(5x - 2)(x + 2)
$$

✔ Answer: $ (5x - 2)(x + 2) $

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2. $ x^2 - 100 $

This is a difference of squares:
$$
x^2 - 100 = x^2 - 10^2 = (x - 10)(x + 10)
$$

✔ Answer: $ (x - 10)(x + 10) $

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3. $ (4a + b)^2 - 25c^2 $

This is a difference of squares again:
$$
(4a + b)^2 - (5c)^2 = [(4a + b) - 5c][(4a + b) + 5c]
$$

Simplify:
$$
(4a + b - 5c)(4a + b + 5c)
$$

✔ Answer: $ (4a + b - 5c)(4a + b + 5c) $

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4. $ 81b^2 - 36 $

Factor out the GCF first: both terms divisible by 9:
$$
81b^2 - 36 = 9(9b^2 - 4)
$$

Now, $ 9b^2 - 4 $ is a difference of squares:
$$
= 9((3b)^2 - 2^2) = 9(3b - 2)(3b + 2)
$$

✔ Answer: $ 9(3b - 2)(3b + 2) $

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5. $ 4a^2 + 16a + 16 $

Factor out GCF: all terms divisible by 4:
$$
4(a^2 + 4a + 4)
$$

Now factor the trinomial:
$$
a^2 + 4a + 4 = (a + 2)^2
$$

So:
$$
4(a + 2)^2
$$

✔ Answer: $ 4(a + 2)^2 $

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6. $ 12x^3 - 31x^2 + 20x $

Factor out the GCF: $ x $
$$
x(12x^2 - 31x + 20)
$$

Now factor the quadratic: $ 12x^2 - 31x + 20 $

Use AC method:
- $ a = 12 $, $ b = -31 $, $ c = 20 $
- $ ac = 12 \cdot 20 = 240 $
- Need two numbers that multiply to 240 and add to $-31$:
→ $-16$ and $-15$ (since $-16 \cdot -15 = 240$, $-16 -15 = -31$)

Split the middle term:
$$
12x^2 - 16x - 15x + 20
$$

Group:
$$
(12x^2 - 16x) - (15x - 20) = 4x(3x - 4) - 5(3x - 4)
$$

Factor:
$$
(4x - 5)(3x - 4)
$$

So overall:
$$
x(4x - 5)(3x - 4)
$$

✔ Answer: $ x(4x - 5)(3x - 4) $

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7. $ x^2 - 9x + 20 $

Find two numbers that multiply to $20$ and add to $-9$:
→ $-4$ and $-5$

So:
$$
x^2 - 9x + 20 = (x - 4)(x - 5)
$$

✔ Answer: $ (x - 4)(x - 5) $

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8. $ (5x - 6)^2 - (4x - 3)^2 $

This is a difference of squares:
$$
A^2 - B^2 = (A - B)(A + B)
$$
where $ A = 5x - 6 $, $ B = 4x - 3 $

So:
$$
[(5x - 6) - (4x - 3)][(5x - 6) + (4x - 3)]
$$

Simplify each:
- First: $ 5x - 6 - 4x + 3 = x - 3 $
- Second: $ 5x - 6 + 4x - 3 = 9x - 9 $

So:
$$
(x - 3)(9x - 9)
$$

Factor out 9 from second term:
$$
(x - 3) \cdot 9(x - 1) = 9(x - 3)(x - 1)
$$

✔ Answer: $ 9(x - 3)(x - 1) $

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9. $ (3x + y)^2 - 9 $

Again, difference of squares:
$$
(3x + y)^2 - 3^2 = [(3x + y) - 3][(3x + y) + 3]
$$

Simplify:
$$
(3x + y - 3)(3x + y + 3)
$$

✔ Answer: $ (3x + y - 3)(3x + y + 3) $

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10. $ x^2 + 10x - 11 $

Find two numbers that multiply to $-11$ and add to $10$:
→ $11$ and $-1$

So:
$$
x^2 + 10x - 11 = (x + 11)(x - 1)
$$

✔ Answer: $ (x + 11)(x - 1) $

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## ✔ Final Answers:

| Expression | Factored Form |
|-----------|----------------|
| $ 5x^2 + 8x - 4 $ | $ (5x - 2)(x + 2) $ |
| $ x^2 - 100 $ | $ (x - 10)(x + 10) $ |
| $ (4a + b)^2 - 25c^2 $ | $ (4a + b - 5c)(4a + b + 5c) $ |
| $ 81b^2 - 36 $ | $ 9(3b - 2)(3b + 2) $ |
| $ 4a^2 + 16a + 16 $ | $ 4(a + 2)^2 $ |
| $ 12x^3 - 31x^2 + 20x $ | $ x(4x - 5)(3x - 4) $ |
| $ x^2 - 9x + 20 $ | $ (x - 4)(x - 5) $ |
| $ (5x - 6)^2 - (4x - 3)^2 $ | $ 9(x - 3)(x - 1) $ |
| $ (3x + y)^2 - 9 $ | $ (3x + y - 3)(3x + y + 3) $ |
| $ x^2 + 10x - 11 $ | $ (x + 11)(x - 1) $ |

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