§ 6.6

Solving Quadratic Equations by Factoring

Martin-Gay, Beginning and Intermediate Algebra, 4ed 22

Zero Factor Theorem

Quadratic Equations• Can be written in the form ax2 + bx + c = 0.• a, b and c are real numbers and a 0.• This is referred to as standard form.

Zero Factor Theorem• If a and b are real numbers and ab = 0, then a = 0 or b = 0.

• This theorem is very useful in solving quadratic equations.

Martin-Gay, Beginning and Intermediate Algebra, 4ed 33

Steps for Solving a Quadratic Equation by Factoring

1) Write the equation in standard form so that one side of the equation is 0.

2) Factor the quadratic expression completely.3) Set each factor containing a variable equal to 0.4) Solve the resulting equations.5) Check each solution in the original equation.

Solving Quadratic Equations

Martin-Gay, Beginning and Intermediate Algebra, 4ed 44

Solve x2 – 5x = 24.• First write the quadratic equation in standard form.

x2 – 5x – 24 = 0• Now we factor the quadratic using techniques from

the previous sections. x2 – 5x – 24 = (x – 8)(x + 3) = 0

• We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = – 3

Solving Quadratic Equations

Continued.

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 55

• Check both possible answers in the original equation.

82 – 5(8) = 64 – 40 = 24 true (–3)2 – 5(–3) = 9 – (–15) = 24 true

• So our solutions for x are 8 or –3.

Solving Quadratic Equations

Example continued:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 66

Solve 4x(8x + 9) = 5• First write the quadratic equation in standard form.

32x2 + 36x = 5 32x2 + 36x – 5 = 0

• Now we factor the quadratic using techniques from the previous sections.

32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0• We set each factor equal to 0.

8x – 1 = 0 or 4x + 5 = 0

Solving Quadratic Equations

Continued.

8x = 1 or 4x = – 5, which simplifies to x = or5.418

Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 77

• Check both possible answers in the original equation.

1 1 14 8 9 4 1 9 4 (10) (10) 581

818 8 2

true

5 54 8 9 4 10 9 4 ( 1) ( 5)( 1) 545 54 44

true

• So our solutions for x are or .81

45

Solving Quadratic Equations

Example continued:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 88

Recall that in Chapter 3, we found the x-intercept of linear equations by letting y = 0 and solving for x. The same method works for x-intercepts in quadratic equations.Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x2 term.The intercepts will be where the parabola crosses the x-axis.

Finding x-intercepts

Martin-Gay, Beginning and Intermediate Algebra, 4ed 99

Find the x-intercepts of the graph of y = 4x2 + 11x + 6.The equation is already written in standard form, so we let y = 0, then factor the quadratic in x.0 = 4x2 + 11x + 6 = (4x + 3)(x + 2)We set each factor equal to 0 and solve for x.4x + 3 = 0 or x + 2 = 04x = –3 or x = –2x = –¾ or x = –2So the x-intercepts are the points (–¾, 0) and (–2, 0).

Finding x-intercepts

Example: