Quadratic by Factoring

Grade 9 · algebra · 71 practice problems · read aloud

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Quadratic Equations by Factoring

This is a method for solving quadratic equations (equations where the highest power of the variable is 2, like x²). It's useful because it turns a complex problem into simpler ones, helping us find where a parabola crosses the x-axis. ✨

Step-by-Step Guide

  1. Set to Zero: Make sure the equation is equal to zero (e.g., ax² + bx + c = 0).
  2. Factor Completely: Factor the quadratic expression on the left side.
  3. Apply Zero Product Property: If (A)(B) = 0, then either A = 0 or B = 0.
  4. Solve Each Mini-Equation: Solve the simple equations from the previous step.
  5. Check Your Solutions: Plug your answers back into the original equation to verify.

Visual Examples

Example 1: Solve x² + 5x + 6 = 0

  1. Factor: (x + 2)(x + 3) = 0
  2. Set each factor to zero: x + 2 = 0 or x + 3 = 0
  3. Solve: x = -2 or x = -3

Example 2: Solve 2x² - 8x = 0

  1. Factor out the GCF (2x): 2x(x - 4) = 0
  2. Set each factor to zero: 2x = 0 or x - 4 = 0
  3. Solve: x = 0 or x = 4

Common Mistakes to Avoid

❌ Forgetting the "= 0": You can only use the Zero Product Property if the product is zero!

❌ Incorrect Factoring: Always double-check your factors by expanding them mentally (FOIL).

❌ Ignoring a Solution: If you have two factors, you should get TWO solutions (they might be the same number).

❌ Dividing by a variable: Never divide both sides by 'x'—you might lose a solution (like x=0 in Example 2). Always factor instead.

Tips & Tricks

Memory Aid: Remember "I.P.S.E.Z" - Isolate (set to zero), Product (factor), Split (set each factor=0), Equate (solve each), Zero (check).

Shortcut for Simple Cases: In x² + bx + c, look for two numbers that multiply to 'c' and add to 'b'.

Strategy: Always look for a Greatest Common Factor (GCF) first—it makes the rest of the factoring much easier!

How to Practice

  • Start with simple equations (like Example 1) to build confidence.
  • Move on to equations that need a GCF factored out first (like Example 2).
  • Challenge yourself with equations where a ≠ 1 (e.g., 2x² + 7x + 3 = 0).
  • Create your own problems, solve them, and then have a friend check your work.
  • Use online practice websites for instant feedback on your answers.

Practice problems

6 of the 71, worked through step by step — try them before opening the answer.

1 x² + 5x = 0

Hint: Factor out the common term from both terms on the left side of the equation. Then apply the zero product property.

Show the answer

Answer: x = 0, -5

  1. Factor the left side of the equation: x(x + 5) = 0
  2. Apply the zero product property: If the product of two factors is zero, then at least one of the factors must be zero.
  3. Set each factor equal to zero and solve: x = 0 x + 5 = 0 → x = -5
  4. The solutions are x = 0 and x = -5.

2 x² - 8x - 33 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term. Since the constant is negative, one factor will be positive and the other negative.

Show the answer

Answer: x = 11, x = -3

  1. Factor the quadratic x² - 8x - 33 = 0.
  2. Find two numbers that multiply to -33 and add to -8. The numbers are -11 and 3 because (-11) × 3 = -33 and (-11) + 3 = -8.
  3. Write the factored form: (x - 11)(x + 3) = 0.
  4. Apply the zero product property: x - 11 = 0 or x + 3 = 0.
  5. Solve each equation: x = 11 or x = -3. The solutions are x = 11 and x = -3.

3 x² - 9x + 14 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term. Since the constant is positive and the middle coefficient is negative, both numbers will be negative.

Show the answer

Answer: x = 2, x = 7

  1. Identify the quadratic equation: x² - 9x + 14 = 0
  2. Find two numbers that multiply to 14 and add to -9. The numbers are -2 and -7 because (-2) × (-7) = 14 and (-2) + (-7) = -9.
  3. Write the factored form: (x - 2)(x - 7) = 0
  4. Apply the zero product property: x - 2 = 0 or x - 7 = 0
  5. Solve each equation: x = 2 or x = 7 The solutions are x = 2 and x = 7.

4 x² - 7x - 18 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term. Since the constant is negative, one factor will be positive and the other negative.

Show the answer

Answer: x = 9, x = -2

  1. Factor the quadratic x² - 7x - 18 = 0.
  2. Find two numbers that multiply to -18 and add to -7. The numbers are -9 and 2 because (-9) × 2 = -18 and (-9) + 2 = -7.
  3. Write the factored form: (x - 9)(x + 2) = 0.
  4. Apply the zero product property: x - 9 = 0 or x + 2 = 0.
  5. Solve each equation: x = 9 or x = -2. The solutions are x = 9 and x = -2.

5 x² - 8x - 48 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term. Since the constant is negative, one factor will be positive and the other negative.

Show the answer

Answer: x = 12, x = -4

  1. Factor the quadratic x² - 8x - 48 = 0.
  2. Find two numbers that multiply to -48 and add to -8. The numbers are -12 and 4 because (-12) × 4 = -48 and (-12) + 4 = -8.
  3. Write the factored form: (x - 12)(x + 4) = 0.
  4. Apply the zero product property: x - 12 = 0 or x + 4 = 0.
  5. Solve each equation: x = 12 or x = -4. The solutions are x = 12 and x = -4.

6 x² + 13x + 42 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term

Show the answer

Answer: x = -6, -7

  1. Factor the quadratic equation x² + 13x + 42 = 0
  2. Find two numbers that multiply to 42 and add to 13
  3. The numbers are 6 and 7 since 6 × 7 = 42 and 6 + 7 = 13
  4. Write the factored form: (x + 6)(x + 7) = 0
  5. Apply the zero product property: x + 6 = 0 or x + 7 = 0
  6. Solve each equation: x = -6 or x = -7
  7. The solutions are x = -6 and x = -7
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