Factoring Quadratic Equations

Factoring is one of the most common methods for solving quadratic equations. The goal is to express the quadratic equation in factored form, which makes it easier to find the solutions. A quadratic equation is typically written as:

$$ ax^2 + bx + c = 0 $$

To factor a quadratic equation, we look for two numbers that multiply to give \( a \times c \) (the product of the leading coefficient and the constant term) and add up to \( b \) (the middle term).

Steps to Factor a Quadratic Equation

  1. Write the equation in standard form: \( ax^2 + bx + c = 0 \).
  2. Identify the values of \( a \), \( b \), and \( c \).
  3. Find two numbers that multiply to \( a \times c \) and add up to \( b \).
  4. Rewrite the middle term using the two numbers from step 3.
  5. Factor by grouping.
  6. Set each factor equal to zero and solve for \( x \).

Example 1: Factoring a Simple Quadratic Equation

Consider the equation:

$$ x^2 + 5x + 6 = 0 $$

Here, \( a = 1 \), \( b = 5 \), and \( c = 6 \).

We need to find two numbers that multiply to \( 1 \times 6 = 6 \) and add up to \( 5 \). These numbers are 2 and 3 because:

  • 2 × 3 = 6
  • 2 + 3 = 5

Now, we can rewrite the middle term as \( 2x + 3x \), so the equation becomes:

$$ x^2 + 2x + 3x + 6 = 0 $$

Next, we factor by grouping:

$$ (x^2 + 2x) + (3x + 6) = 0 $$

Now, factor out the greatest common factor (GCF) from each group:

$$ x(x + 2) + 3(x + 2) = 0 $$

Finally, factor out the common binomial \( (x + 2) \):

$$ (x + 2)(x + 3) = 0 $$

Now, set each factor equal to zero and solve for \( x \):

  • \( x + 2 = 0 \Rightarrow x = -2 \)
  • \( x + 3 = 0 \Rightarrow x = -3 \)

The solutions are \( x = -2 \) and \( x = -3 \).

Example 2: Factoring a More Complex Quadratic Equation

Consider the equation:

$$ 2x^2 + 7x + 3 = 0 $$

Here, \( a = 2 \), \( b = 7 \), and \( c = 3 \).

We need to find two numbers that multiply to \( 2 \times 3 = 6 \) and add up to \( 7 \). These numbers are 1 and 6 because:

  • 1 × 6 = 6
  • 1 + 6 = 7

Now, we can rewrite the middle term as \( 1x + 6x \), so the equation becomes:

$$ 2x^2 + x + 6x + 3 = 0 $$

Next, we factor by grouping:

$$ (2x^2 + x) + (6x + 3) = 0 $$

Now, factor out the GCF from each group:

$$ x(2x + 1) + 3(2x + 1) = 0 $$

Finally, factor out the common binomial \( (2x + 1) \):

$$ (2x + 1)(x + 3) = 0 $$

Now, set each factor equal to zero and solve for \( x \):

  • \( 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \)
  • \( x + 3 = 0 \Rightarrow x = -3 \)

The solutions are \( x = -\frac{1}{2} \) and \( x = -3 \).

Practice Questions

  1. Factor the equation \( x^2 + 7x + 10 = 0 \).
    Solution

    Identify \( a = 1 \), \( b = 7 \), and \( c = 10 \).

    We need to find two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5.

    $$ (x + 2)(x + 5) = 0 $$

    • \( x = -2 \)
    • \( x = -5 \)

    The solutions are \( x = -2 \) and \( x = -5 \).

  2. Factor the equation \( 3x^2 + 5x - 2 = 0 \).
    Solution

    Identify \( a = 3 \), \( b = 5 \), and \( c = -2 \).

    We need to find two numbers that multiply to \( 3 \times -2 = -6 \) and add up to 5. These numbers are 6 and -1.

    $$ 3x^2 + 6x - x - 2 = 0 $$

    Now, group and factor:

    $$ 3x(x + 2) - 1(x + 2) = 0 $$

    Factor out the common binomial:

    $$ (3x - 1)(x + 2) = 0 $$

    • \( 3x - 1 = 0 \Rightarrow x = \frac{1}{3} \)
    • \( x + 2 = 0 \Rightarrow x = -2 \)

    The solutions are \( x = \frac{1}{3} \) and \( x = -2 \).

  3. Factor the equation \( x^2 - 9x + 20 = 0 \).
    Solution

    Identify \( a = 1 \), \( b = -9 \), and \( c = 20 \).

    We need to find two numbers that multiply to 20 and add up to -9. These numbers are -4 and -5.

    $$ (x - 4)(x - 5) = 0 $$

    • \( x = 4 \)
    • \( x = 5 \)

    The solutions are \( x = 4 \) and \( x = 5 \).

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