Solving Quadratic Equationsby Factoring

A quadratic equation is one that can be written in the form …

Ax² + bx + C = 0,

Where a, b, and c are real numbers and A does not equal zero!

Examples: x² + 3x + 4 = 0, y² + y = 1, x² = 16

Standard Form: Ax² + bx + C = 0

Zero Factor TheoremIf A and B are real numbers and if

AB = 0, then A = 0 or B = 0

Example: (x + 2)(x+ 3) = 0

Then, x + 2 = 0 and x + 3 = 0 x = -2 x = -3

We can only use this theorem if the product(AB) = 0

• Example: (3x-8)(x – 15) = 0

3x – 8 = 0 x – 15 = 03x = 8 x = 15X = 8/3

X(2x – 5) = 0

X = 0 2x – 5 = 0 2x = 5x = 5/2

Solving Quadratic Equations by Factoring

• Write the equation in standard form Ax² + bx + C = 0

• Factor the quadratic completely!• Set each factor equal to zero and solve• Check your answer

X² - 19x – 20 = 0

First we need to factor!

(x – 20)(x + 1) = 0

X – 20 = 0 x + 1 = 0X = 20 x = -1

Example: 2x² -7x + 5 = 0 2x² -2x – 5x + 5 = 02x(x – 1) + -5(x – 1) = 0(x -1)(2x – 5) = 0X = 1 x = 5/2

Practice

• X ² - 14x + 24 = 0

• X² - 36 = 0

• 9x² - 4 = 0

• 9x² + 6x + 2 = 0

• (x – 12)(x-2) = 0 x = 2, 12

• (x+6)(x-6) = 0 x = -6, 6

• (3x – 2)(3x + 2) x = -2/3, 2/3

• Prime polynomial!No real solution!

What happens when it doesn’t equal zero?

X² - 5x = 14 -14 -14

X² - 5x -14 = 0

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

• X(2x – 7) = 4Should I set everything equal to 4?• No! You need to write your equation in

standard form!X(2x – 7) – 4 = 0

2x ² - 7x – 4 = 02x ² - 8x + 1x – 4 = 02x(x – 4) + 1 (x -4) = 0(x-4)(2x + 1) = 0X – 4 = 0 2x + 1 = 0X = 4 x = -1/2

-5x² + 60 = -20x-5x² + 20x + 60-5(x² - 4x - 12) = 0

-5(x – 6)(x+ 2) = 0

-5 = 0

X – 6 = 0

X + 2 = 0

X = 6, -2

3x² = 12x

3x² - 12x = 0

3x(x – 4) = 0

3x = 0

X – 4 = 0

X = 0, 4

Practice

• X² - 13x = -36

• X² = 16

• X² + 12x = -36

X² - 13x + 36 = 0(x – 9)(x -4) = 0X = 9, 4

X² - 16 = 0 (x-4)(x +4) = 0X = -4, 4

X² + 12x + 36 = 0(x +6)²X + 6 = 0X = -6

Polynomials Higher than 2nd degree

2x³ - 18x = 0GCF: 2x

2x(x² - 9) = 0

2x(x-3)(x+ 3) = 02x = 0 x – 3 = 0 x + 3 = 0X = 0, 3, -3

Practice

• (x+ 3)(3x² - 20x – 7) = 0

• 3x³ - 9x² - 12x = 0

(x + 3)(3x² - 21x + 1x – 7) = 0 (x+3)(3x(x-7) + 1(x-7)) = 0(x + 3)(x-7)(3x + 1) = 0X + 3 = 0; x – 7 = 0; 3x + 1 = 0X = -3, 7, -1/3

3x(x² - 3x – 4) = 03x(x – 4)(x + 1) = 03x = 0 x- 4 = 0 x + 1 = 0X = 0, 4, -1