Solving Quadratic Equations by Factoring Matching Worksheet by ... - Free Printable

Problem Overview:

The task involves solving quadratic equations by factoring and using the Zero Product Property. The Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. This property is used to solve equations where the quadratic expression is set equal to zero.

Let's solve each problem step by step.

---

Part 1: Using the Zero Product Property

#### Problem 1:
\[
(x - 1)(x - 5) = 0
\]
- According to the Zero Product Property, either:
\[
x - 1 = 0 \quad \text{or} \quad x - 5 = 0
\]
- Solving these:
\[
x - 1 = 0 \implies x = 1
\]
\[
x - 5 = 0 \implies x = 5
\]
- Solutions: \( x = 1 \) or \( x = 5 \)

#### Problem 2:
\[
(x - 2)(x - 9) = 0
\]
- According to the Zero Product Property, either:
\[
x - 2 = 0 \quad \text{or} \quad x - 9 = 0
\]
- Solving these:
\[
x - 2 = 0 \implies x = 2
\]
\[
x - 9 = 0 \implies x = 9
\]
- Solutions: \( x = 2 \) or \( x = 9 \)

#### Problem 3:
\[
(x - 2)(x + 4) = 0
\]
- According to the Zero Product Property, either:
\[
x - 2 = 0 \quad \text{or} \quad x + 4 = 0
\]
- Solving these:
\[
x - 2 = 0 \implies x = 2
\]
\[
x + 4 = 0 \implies x = -4
\]
- Solutions: \( x = 2 \) or \( x = -4 \)

#### Problem 4:
\[
(2x + 1)(x - 6) = 0
\]
- According to the Zero Product Property, either:
\[
2x + 1 = 0 \quad \text{or} \quad x - 6 = 0
\]
- Solving these:
\[
2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}
\]
\[
x - 6 = 0 \implies x = 6
\]
- Solutions: \( x = -\frac{1}{2} \) or \( x = 6 \)

---

Part 2: Solving Quadratic Equations by Factoring

#### Problem 5:
\[
x^2 - 3x = 0
\]
- Factor out the greatest common factor (GCF):
\[
x(x - 3) = 0
\]
- According to the Zero Product Property, either:
\[
x = 0 \quad \text{or} \quad x - 3 = 0
\]
- Solving these:
\[
x = 0
\]
\[
x - 3 = 0 \implies x = 3
\]
- Solutions: \( x = 0 \) or \( x = 3 \)

#### Problem 6:
\[
x^2 + 4x + 3 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 4x + 3 = (x + 1)(x + 3)
\]
- According to the Zero Product Property, either:
\[
x + 1 = 0 \quad \text{or} \quad x + 3 = 0
\]
- Solving these:
\[
x + 1 = 0 \implies x = -1
\]
\[
x + 3 = 0 \implies x = -3
\]
- Solutions: \( x = -1 \) or \( x = -3 \)

#### Problem 7:
\[
x^2 + 5x - 6 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 5x - 6 = (x + 6)(x - 1)
\]
- According to the Zero Product Property, either:
\[
x + 6 = 0 \quad \text{or} \quad x - 1 = 0
\]
- Solving these:
\[
x + 6 = 0 \implies x = -6
\]
\[
x - 1 = 0 \implies x = 1
\]
- Solutions: \( x = -6 \) or \( x = 1 \)

#### Problem 8:
\[
x^2 + 11x + 24 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 11x + 24 = (x + 3)(x + 8)
\]
- According to the Zero Product Property, either:
\[
x + 3 = 0 \quad \text{or} \quad x + 8 = 0
\]
- Solving these:
\[
x + 3 = 0 \implies x = -3
\]
\[
x + 8 = 0 \implies x = -8
\]
- Solutions: \( x = -3 \) or \( x = -8 \)

#### Problem 9:
\[
x^2 - 12x + 11 = 0
\]
- Factor the quadratic expression:
\[
x^2 - 12x + 11 = (x - 1)(x - 11)
\]
- According to the Zero Product Property, either:
\[
x - 1 = 0 \quad \text{or} \quad x - 11 = 0
\]
- Solving these:
\[
x - 1 = 0 \implies x = 1
\]
\[
x - 11 = 0 \implies x = 11
\]
- Solutions: \( x = 1 \) or \( x = 11 \)

#### Problem 10:
\[
x^2 + 18x - 65 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 18x - 65 = (x + 13)(x - 5)
\]
- According to the Zero Product Property, either:
\[
x + 13 = 0 \quad \text{or} \quad x - 5 = 0
\]
- Solving these:
\[
x + 13 = 0 \implies x = -13
\]
\[
x - 5 = 0 \implies x = 5
\]
- Solutions: \( x = -13 \) or \( x = 5 \)

#### Problem 11:
\[
x^2 - 4x - 12 = 0
\]
- Factor the quadratic expression:
\[
x^2 - 4x - 12 = (x - 6)(x + 2)
\]
- According to the Zero Product Property, either:
\[
x - 6 = 0 \quad \text{or} \quad x + 2 = 0
\]
- Solving these:
\[
x - 6 = 0 \implies x = 6
\]
\[
x + 2 = 0 \implies x = -2
\]
- Solutions: \( x = 6 \) or \( x = -2 \)

#### Problem 12:
\[
x^2 + 11x + 10 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 11x + 10 = (x + 1)(x + 10)
\]
- According to the Zero Product Property, either:
\[
x + 1 = 0 \quad \text{or} \quad x + 10 = 0
\]
- Solving these:
\[
x + 1 = 0 \implies x = -1
\]
\[
x + 10 = 0 \implies x = -10
\]
- Solutions: \( x = -1 \) or \( x = -10 \)

#### Problem 13:
\[
x^2 + 12x + 35 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 12x + 35 = (x + 5)(x + 7)
\]
- According to the Zero Product Property, either:
\[
x + 5 = 0 \quad \text{or} \quad x + 7 = 0
\]
- Solving these:
\[
x + 5 = 0 \implies x = -5
\]
\[
x + 7 = 0 \implies x = -7
\]
- Solutions: \( x = -5 \) or \( x = -7 \)

#### Problem 14:
\[
2x^2 - 3x - 5 = 0
\]
- Factor the quadratic expression:
\[
2x^2 - 3x - 5 = (2x - 5)(x + 1)
\]
- According to the Zero Product Property, either:
\[
2x - 5 = 0 \quad \text{or} \quad x + 1 = 0
\]
- Solving these:
\[
2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2}
\]
\[
x + 1 = 0 \implies x = -1
\]
- Solutions: \( x = \frac{5}{2} \) or \( x = -1 \)

#### Problem 15:
\[
3x^2 - 5x - 2 = 0
\]
- Factor the quadratic expression:
\[
3x^2 - 5x - 2 = (3x + 1)(x - 2)
\]
- According to the Zero Product Property, either:
\[
3x + 1 = 0 \quad \text{or} \quad x - 2 = 0
\]
- Solving these:
\[
3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}
\]
\[
x - 2 = 0 \implies x = 2
\]
- Solutions: \( x = -\frac{1}{3} \) or \( x = 2 \)

#### Problem 16:
\[
x^2 = 3x + 40
\]
- Rewrite the equation in standard form:
\[
x^2 - 3x - 40 = 0
\]
- Factor the quadratic expression:
\[
x^2 - 3x - 40 = (x - 8)(x + 5)
\]
- According to the Zero Product Property, either:
\[
x - 8 = 0 \quad \text{or} \quad x + 5 = 0
\]
- Solving these:
\[
x - 8 = 0 \implies x = 8
\]
\[
x + 5 = 0 \implies x = -5
\]
- Solutions: \( x = 8 \) or \( x = -5 \)

#### Problem 17:
\[
x^2 - 14 = -5x
\]
- Rewrite the equation in standard form:
\[
x^2 + 5x - 14 = 0
\]
- Factor the quadratic expression:
\[
x^2 + 5x - 14 = (x + 7)(x - 2)
\]
- According to the Zero Product Property, either:
\[
x + 7 = 0 \quad \text{or} \quad x - 2 = 0
\]
- Solving these:
\[
x + 7 = 0 \implies x = -7
\]
\[
x - 2 = 0 \implies x = 2
\]
- Solutions: \( x = -7 \) or \( x = 2 \)

#### Problem 18:
\[
2x - 1 = -8x^2
\]
- Rewrite the equation in standard form:
\[
8x^2 + 2x - 1 = 0
\]
- Factor the quadratic expression:
\[
8x^2 + 2x - 1 = (4x - 1)(2x + 1)
\]
- According to the Zero Product Property, either:
\[
4x - 1 = 0 \quad \text{or} \quad 2x + 1 = 0
\]
- Solving these:
\[
4x - 1 = 0 \implies 4x = 1 \implies x = \frac{1}{4}
\]
\[
2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}
\]
- Solutions: \( x = \frac{1}{4} \) or \( x = -\frac{1}{2} \)

#### Problem 19:
\[
x^2 = 10x^2 - 2
\]
- Rewrite the equation in standard form:
\[
x^2 - 10x^2 + 2 = 0 \implies -9x^2 + 2 = 0
\]
- Simplify:
\[
-9x^2 + 2 = 0 \implies 9x^2 = 2 \implies x^2 = \frac{2}{9}
\]
- Solve for \( x \):
\[
x = \pm \sqrt{\frac{2}{9}} = \pm \frac{\sqrt{2}}{3}
\]
- Solutions: \( x = \frac{\sqrt{2}}{3} \) or \( x = -\frac{\sqrt{2}}{3} \)

#### Problem 20:
\[
2x^2 = 13x + 7
\]
- Rewrite the equation in standard form:
\[
2x^2 - 13x - 7 = 0
\]
- Factor the quadratic expression:
\[
2x^2 - 13x - 7 = (2x + 1)(x - 7)
\]
- According to the Zero Product Property, either:
\[
2x + 1 = 0 \quad \text{or} \quad x - 7 = 0
\]
- Solving these:
\[
2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}
\]
\[
x - 7 = 0 \implies x = 7
\]
- Solutions: \( x = -\frac{1}{2} \) or \( x = 7 \)

#### Problem 21:
\[
6x^2 + x = 5
\]
- Rewrite the equation in standard form:
\[
6x^2 + x - 5 = 0
\]
- Factor the quadratic expression:
\[
6x^2 + x - 5 = (3x + 5)(2x - 1)
\]
- According to the Zero Product Property, either:
\[
3x + 5 = 0 \quad \text{or} \quad 2x - 1 = 0
\]
- Solving these:
\[
3x + 5 = 0 \implies 3x = -5 \implies x = -\frac{5}{3}
\]
\[
2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}
\]
- Solutions: \( x = -\frac{5}{3} \) or \( x = \frac{1}{2} \)

#### Problem 22:
\[
x^2 = 5x
\]
- Rewrite the equation in standard form:
\[
x^2 - 5x = 0
\]
- Factor out the GCF:
\[
x(x - 5) = 0
\]
- According to the Zero Product Property, either:
\[
x = 0 \quad \text{or} \quad x - 5 = 0
\]
- Solving these:
\[
x = 0
\]
\[
x - 5 = 0 \implies x = 5
\]
- Solutions: \( x = 0 \) or \( x = 5 \)

---

Part 3: Application Problem

#### Problem 23:
The height of a flare fired from the deck of a ship is modeled by:
\[
h = -16t^2 + 104t + 56
\]
where \( h \) is the height above water and \( t \) is the time in seconds. Find the time it takes the flare to hit the water.

- When the flare hits the water, the height \( h \) is 0:
\[
-16t^2 + 104t + 56 = 0
\]
- Simplify the equation by dividing through by -8:
\[
2t^2 - 13t - 7 = 0
\]
- Factor the quadratic expression:
\[
2t^2 - 13t - 7 = (2t + 1)(t - 7)
\]
- According to the Zero Product Property, either:
\[
2t + 1 = 0 \quad \text{or} \quad t - 7 = 0
\]
- Solving these:
\[
2t + 1 = 0 \implies 2t = -1 \implies t = -\frac{1}{2}
\]
\[
t - 7 = 0 \implies t = 7
\]
- Since time cannot be negative, we discard \( t = -\frac{1}{2} \).

- Solution: \( t = 7 \)

---

Final Answers:

\[
\boxed{
\begin{aligned}
&1. \, x = 1 \, \text{or} \, x = 5 \\
&2. \, x = 2 \, \text{or} \, x = 9 \\
&3. \, x = 2 \, \text{or} \, x = -4 \\
&4. \, x = -\frac{1}{2} \, \text{or} \, x = 6 \\
&5. \, x = 0 \, \text{or} \, x = 3 \\
&6. \, x = -1 \, \text{or} \, x = -3 \\
&7. \, x = -6 \, \text{or} \, x = 1 \\
&8. \, x = -3 \, \text{or} \, x = -8 \\
&9. \, x = 1 \, \text{or} \, x = 11 \\
&10. \, x = -13 \, \text{or} \, x = 5 \\
&11. \, x = 6 \, \text{or} \, x = -2 \\
&12. \, x = -1 \, \text{or} \, x = -10 \\
&13. \, x = -5 \, \text{or} \, x = -7 \\
&14. \, x = \frac{5}{2} \, \text{or} \, x = -1 \\
&15. \, x = -\frac{1}{3} \, \text{or} \, x = 2 \\
&16. \, x = 8 \, \text{or} \, x = -5 \\
&17. \, x = -7 \, \text{or} \, x = 2 \\
&18. \, x = \frac{1}{4} \, \text{or} \, x = -\frac{1}{2} \\
&19. \, x = \frac{\sqrt{2}}{3} \, \text{or} \, x = -\frac{\sqrt{2}}{3} \\
&20. \, x = -\frac{1}{2} \, \text{or} \, x = 7 \\
&21. \, x = -\frac{5}{3} \, \text{or} \, x = \frac{1}{2} \\
&22. \, x = 0 \, \text{or} \, x = 5 \\
&23. \, t = 7
\end{aligned}
}
\]