a method for breaking up a quadratic equation in the form ax 2 + bx + c into factors (expressions...

Factoring

Factoring Quadratic Trinomials A method for breaking up a

quadratic equation in the form ax2 + bx + c into factors (expressions which multiply to give you the original trinomial).

Use Algebra tiles to help you develop the patterns in factoring.

Example Set 1

x2 + 5x + 4

x2 + 7x + 12

x2 + 6x + 8

Example Set 2

x2 – 6x + 8

x2 - 9x + 20

x2 – 4x + 4

Notice how the answer changes if the b term is negative.

Example Set 3

x2 + 4x – 12

x2 + 3x – 18

x2 – 7x – 8

x2 – 3x – 10

Mixed Practice

Factor each of the following expressions.

x2 + 14x + 48

x2 – 22x – 48

x2 – 13x + 42

x2 + 5x - 24

Other factoring methods

Difference of squares a2 – b2 = (a + b)(a – b)

Examples x2 – 49 = (x + 7)(x – 7) 4x2 – 1 = (2x + 1)(2x – 1)

Practice Factor: x2 – 64 Factor: 9x2 - 16

Factoring Out a Common Monomial Check to see if each term in the

expression has a common factor. Examples:

8x3 – 50x What factors do they have in common?

2x(4x2 – 25) 2x(2x + 5)(2x – 5)

Factor the following 5x4 – 10x3

2x2 + 28x + 96

Steps in Factoring

1. Try to factor out a common monomial factor from all terms first.

2. Then look to see if the difference of squares format exists.

3. Use your techniques for factoring a quadratic trinomial if applicable.

Do-Now: Factor the following…. 2x2 – 18x + 40 3x3 – 9x2 – 30x

Factoring when a > 1

What makes this expression more difficult to factor than what we have done before? 2x2 + 7x + 6

You can use guess and check using FOIL, but it can become tedious for more difficult problems. Try these using guess and check. 3x2 – 10x + 8 6x2 + x – 15

The A-C method of factoring. You may want to use this method on more

complicated examples, such as…….. 6x2 + x – 12

First identify the values of a and c.▪ a = 6, c = -12

Next, multiply a times c.▪ (6)(-12) = -72

Then find two numbers that multiply to be (a)(c) and add to be the value of b.▪ (9)(-8) = -72, 9 + -8 = 1

A-C method continued

Now rewrite the original expression with the x-term written as the sum of the two numbers you found. 6x2 + x – 12▪ 6x2 – 8x + 9x – 12

Finally, break the expression into two parts and factor twice. (6x2 – 8x) + (9x – 12) 2x(3x – 4) + 3(3x – 4) (3x – 4)(2x + 3)

Factor the following

2x2 + 5x – 3

9x2 + 12x + 4

12x2 + 17x + 6

15x2 + 8x – 16

20x2 – 54x + 36

How does factoring help? It allows us to solve: ax2 + bx + c =

0 using the “Zero Product Property” If the product of two expressions is

equal to zero, then……… one or both of the expressions must

equal zero. Examples. Solve the following

equations. x2 – x – 42 = 0 x2 = 7x

Additional Practice

Factor: 3x2 + 17x + 10 8x2 + 4x – 24

Solve the following equation. 4x2 + 10x – 24 = 0

Roots/zeros of a function

When you solve y = ax2 + bx + c when y = 0, you are finding the roots or zeros of the function. These are the points where the parabola crosses the x-axis.