Math worksheet for solving quadratic equations in factored form.

Let’s solve each problem step by step. We’ll use the Zero-Product Property, which says:

> If two things multiply to give zero, then at least one of them must be zero.

So if we have something like:
(a)(b) = 0, then either a = 0 or b = 0 (or both).

We’ll apply this to each equation.

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Problem 1: Write the Zero-Product Property

The Zero-Product Property states:
If the product of two factors is zero, then at least one of the factors must be zero.
In math terms: If \( a \cdot b = 0 \), then \( a = 0 \) or \( b = 0 \).

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Now let’s solve for x in each problem.

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Problem 2: \( (x - 3)(x + 4) = 0 \)

Set each factor equal to zero:

→ \( x - 3 = 0 \) → \( x = 3 \)
→ \( x + 4 = 0 \) → \( x = -4 \)

✔ Solutions: \( x = 3 \) or \( x = -4 \)

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Problem 3: \( (2x - 4)(x + 5) = 0 \)

Set each factor to zero:

→ \( 2x - 4 = 0 \) → Add 4: \( 2x = 4 \) → Divide by 2: \( x = 2 \)
→ \( x + 5 = 0 \) → \( x = -5 \)

✔ Solutions: \( x = 2 \) or \( x = -5 \)

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Problem 4: \( x(2x - 4) = 0 \)

This is already factored. Set each part to zero:

→ \( x = 0 \)
→ \( 2x - 4 = 0 \) → \( 2x = 4 \) → \( x = 2 \)

✔ Solutions: \( x = 0 \) or \( x = 2 \)

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Problem 5: \( (x + 3)(4x + 1) = 0 \)

Set each factor to zero:

→ \( x + 3 = 0 \) → \( x = -3 \)
→ \( 4x + 1 = 0 \) → Subtract 1: \( 4x = -1 \) → Divide by 4: \( x = -\frac{1}{4} \)

✔ Solutions: \( x = -3 \) or \( x = -\frac{1}{4} \)

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Problem 6: \( 12(3x + 6)(x - 2) = 0 \)

Note: The 12 is just a number — it doesn’t affect the solutions because 12 ≠ 0. So we ignore it and set the factors to zero:

→ \( 3x + 6 = 0 \) → Subtract 6: \( 3x = -6 \) → Divide by 3: \( x = -2 \)
→ \( x - 2 = 0 \) → \( x = 2 \)

✔ Solutions: \( x = -2 \) or \( x = 2 \)

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Problem 7: \( (4x - 9)(x - 2) = 0 \)

Set each factor to zero:

→ \( 4x - 9 = 0 \) → Add 9: \( 4x = 9 \) → Divide by 4: \( x = \frac{9}{4} \)
→ \( x - 2 = 0 \) → \( x = 2 \)

✔ Solutions: \( x = \frac{9}{4} \) or \( x = 2 \)

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Problem 8: \( (x - 4)(3x - 2) = 0 \)

Set each factor to zero:

→ \( x - 4 = 0 \) → \( x = 4 \)
→ \( 3x - 2 = 0 \) → Add 2: \( 3x = 2 \) → Divide by 3: \( x = \frac{2}{3} \)

✔ Solutions: \( x = 4 \) or \( x = \frac{2}{3} \)

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Problem 9: \( (5x + 5)^2 = 0 \)

This means \( (5x + 5)(5x + 5) = 0 \), so same factor twice.

Set it to zero:

→ \( 5x + 5 = 0 \) → Subtract 5: \( 5x = -5 \) → Divide by 5: \( x = -1 \)

Since it’s squared, it’s still only one solution (but repeated).

✔ Solution: \( x = -1 \)

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Problem 10: \( 5(x + 7)^2 = 0 \)

Again, 5 ≠ 0, so focus on \( (x + 7)^2 = 0 \)

→ \( x + 7 = 0 \) → \( x = -7 \)

✔ Solution: \( x = -7 \)

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Problem 11: \( (x - 2)(x + 5) = 0 \)

Set each factor to zero:

→ \( x - 2 = 0 \) → \( x = 2 \)
→ \( x + 5 = 0 \) → \( x = -5 \)

✔ Solutions: \( x = 2 \) or \( x = -5 \)

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Problem 12: \( (x - 3)^2 = 0 \)

Same as before — square means same factor twice.

→ \( x - 3 = 0 \) → \( x = 3 \)

✔ Solution: \( x = 3 \)

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Problem 13: \( (x - 3)(5x + 9) = 0 \)

Set each factor to zero:

→ \( x - 3 = 0 \) → \( x = 3 \)
→ \( 5x + 9 = 0 \) → Subtract 9: \( 5x = -9 \) → Divide by 5: \( x = -\frac{9}{5} \)

✔ Solutions: \( x = 3 \) or \( x = -\frac{9}{5} \)

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Final Answer:

1) Zero-Product Property: If \( a \cdot b = 0 \), then \( a = 0 \) or \( b = 0 \).
2) \( x = 3 \) or \( x = -4 \)
3) \( x = 2 \) or \( x = -5 \)
4) \( x = 0 \) or \( x = 2 \)
5) \( x = -3 \) or \( x = -\frac{1}{4} \)
6) \( x = -2 \) or \( x = 2 \)
7) \( x = \frac{9}{4} \) or \( x = 2 \)
8) \( x = 4 \) or \( x = \frac{2}{3} \)
9) \( x = -1 \)
10) \( x = -7 \)
11) \( x = 2 \) or \( x = -5 \)
12) \( x = 3 \)
13) \( x = 3 \) or \( x = -\frac{9}{5} \)