Solve Quadratic Equations by Factoring

This is a tutorial on how to solve quadratic equations by factoring. There is also Factor Quadratic Expressions - Step by Step Calculator

Example 1 : Solve the following quadratic equation.


x 2 - 3x = 0

Solution to Example 1:

  • Given
    x 2 - 3x = 0

  • Factor x out in the expression on the left.
    x (x - 3) = 0

  • For the product x (x - 3) to be equal to zero we nedd to have
    x = 0 or x - 3 = 0

  • Solve the above simple equations to obtain the solutions.
    x = 0

    or

    x = 3

  • As an exercise, check that x = 0 and x = 3 are solutions to the given equation.

Example 2 : Solve the quadratic equation given below


x 2 - 5 x + 6 = 0

Solution to Example 2:

  • To factor the expression on the left, we need to write x 2 - 5 x + 6 in the form factored:

    x 2 - 5 x + 6 = (x + a)(x + b)

  • so that the sum of a and b is -5 and their product is 6. The numbers that satisfy these conditions are - 2 and - 3. Hence
    x 2 - 5 x + 6 = (x - 2)(x - 3)

  • Substitute into the original equation and solve.
    (x - 2)(x - 3) = 0

  • (x - 2)(x - 3) is equal to zero if
    x - 2 = 0

    or

    x - 3 = 0

  • Solve the above equations to obtain two solutions to the given equation.

    x = 2

    or

    x = 3

  • As an exercise, check that x = 0 and x = 3 are solutions to the given equation.

Example 3: Solve the following equation


2 x 2 + x - 21 = 0

Solution to Example 3:

  • We first try to write 2 x 2 + x - 21 in the factored form
    2 x 2 + x - 21 = (2x + a)(x + b)

  • Such that the product a b is equat to - 21 and a + 2 b = 1

    two pairs of numbers gives a product of - 21: either -3 and 7 or 3 and -7. After some trial exercises it found that 2 x 2 + x - 21 may be factored as follows:

    2 x 2 + x - 21 = (2x + 7)(x - 3)

  • We now substitute into the original equation

    (2x + 7)(x - 3) = 0

  • and solve the following simpler equations
    2x + 7 = 0

    x - 3 = 0

  • to obtain
    x = - 7 / 2

    or x = 3

  • As an exercise, check that x = 0 and x = 3 are solutions to the given equation.

Example 4: Solve the following equation


(x - 1)(x + 1 / 2) = - x + 1

Solution to Example 4:

  • At first we might be tempted into expanding the left side of the equation. However after examination of the right side, the above equation may be written as:
    (x - 1)(x + 1 / 2) = - (x - 1)

  • Write the equation with the right side equal to zero.

    (x - 1)(x + 1 / 2) + (x - 1) = 0

  • We now factor (x - 1) out.

    (x - 1)(x + 1 / 2 + 1) = 0

  • and solve the following simpler equations
    x - 1 = 0

    x + 3 / 2 = 0

  • to obtain
    x = 1

    or

    x = - 3 / 2

More references and links to quadratic equations.