§ 3.6

Solving Quadratic Equations by Factoring

Martin-Gay, Developmental Mathematics 2

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 m and n are real numbers and mn = 0, then m = 0 or n = 0.

• This theorem is very useful in solving quadratic equations.

Martin-Gay, Developmental Mathematics 3

Solving Quadratic Equations

Steps for Solving a Quadratic Equation by Factoring

1) Write the equation in standard form.

2) Factor the quadratic completely.

3) Set each factor containing a variable equal to 0.

4) Solve the resulting equations.

5) Check each solution in the original equation.

Martin-Gay, Developmental Mathematics 4

Solving Quadratic Equations

In general:

ax2 + bx + c = 0.0))(( qnxpmx

0 0 qnxorpmx

n

qx

m

px or

Martin-Gay, Developmental Mathematics 5

Solving Quadratic Equations

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

Example 1

Continued.

Martin-Gay, Developmental Mathematics 6

Solving Quadratic Equations

• 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.

Example Continued

Martin-Gay, Developmental Mathematics 7

Solving Quadratic Equations

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

Example 2

Continued.

8x = 1 or 4x = – 5, which simplifies to x = or 5.4

18

Martin-Gay, Developmental Mathematics 8

Solving Quadratic Equations

Put it all together:

4x(8x + 9) = 5

32x2 + 36x = 5

32x2 + 36x – 5 = 0

(8x – 1)(4x + 5) = 0

8x – 1 = 0 or 4x + 5 = 0

Martin-Gay, Developmental Mathematics 9

Solving Quadratic Equations

• Check both possible answers in the original equation.

1 1 14 8 9 4 1 9 4 (10) (10) 58

18

18 8 2

true

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

5 54 44

true

• So our solutions for x are or .8

1

4

5

Example Continued