# Solving Quadratic Equations by Factoring

## Presentation on theme: "Solving Quadratic Equations by Factoring"— Presentation transcript:

Solving Quadratic Equations by 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 Theorem If 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 = 0 3x = x = 15 X = 8/3

X(2x – 5) = 0 X = 0 2x – 5 = 0 2x = 5 x = 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
X² - 19x – 20 = 0 First we need to factor! (x – 20)(x + 1) = 0 X – 20 = 0 x + 1 = 0 X = 20 x = -1 Example: 2x² -7x + 5 = 0 2x² -2x – 5x + 5 = 0 2x(x – 1) + -5(x – 1) = 0 (x -1)(2x – 5) = 0 X = 1 x = 5/2

Practice X ² - 14x + 24 = 0 X² - 36 = 0 9x² - 4 = 0 9x² + 6x + 2 = 0
Prime polynomial! No real solution!

What happens when it doesn’t equal zero?
X² - 5x = X² - 5x -14 = 0 (x + 2)(x-7) = 0 X = -2 x = 7

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

-5x² + 60 = -20x -5x² + 20x (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 = 0 (x – 9)(x -4) = 0 X = 9, 4 X² - 16 = 0 (x-4)(x +4) = 0 X = -4, 4 X² + 12x + 36 = 0 (x +6)² X + 6 = 0 X = -6 X² - 13x = -36 X² = 16 X² + 12x = -36

Polynomials Higher than 2nd degree
2x³ - 18x = 0 GCF: 2x 2x(x² - 9) = 0 2x(x-3)(x+ 3) = 0 2x = 0 x – 3 = 0 x + 3 = 0 X = 0, 3, -3

(x + 3)(3x² - 21x + 1x – 7) = 0 (x+3)(3x(x-7) + 1(x-7)) = 0 (x + 3)(x-7)(3x + 1) = 0 X + 3 = 0; x – 7 = 0; 3x + 1 = 0 X = -3, 7, -1/3 3x(x² - 3x – 4) = 0 3x(x – 4)(x + 1) = 0 3x = 0 x- 4 = 0 x + 1 = 0 X = 0, 4, -1 Practice (x+ 3)(3x² - 20x – 7) = 0 3x³ - 9x² - 12x = 0