Student will be able Student will be able to factor Quadratic to factor Quadratic

Trinomials of the form Trinomials of the form

Student will be able Student will be able to factor Quadratic to factor Quadratic

Trinomials of the form Trinomials of the form

Leading coefficient not = 1Leading coefficient not = 1

2ax bx c

Multiplying Binomials

• Recall that when you multiplied two binomials the result was usually a quadratic trinomial. Foe example,

(2 3)( 1) Using the FOIL Methodx x 2First: 2x

Outer: 2xInner: 3xLast: 3

Combining similar terms, we get

22 3x x

Factoring Trinomials“A special method”

There is a method of factoring quadratic trinomials that ALWAYS works.

When using this method, if it does not work, then the trinomial is PRIME. This means it CANNOT BE FACTORED.

Example #12Factor: 6 8x x

The signs of the trinomial tell you the signs of each binomial

“+” sign, then the factors are the same sign

Tells the signs of each factor

(x +4)(x+2)To get the number factors we look for the factors of 8 (constant term) that ADD up to 6 (middle term)

8 = 4 X 2

4 + 2 = 6

Example #22Factor 5 6x x

“-” sign telss you that the signs of the factors are different

“-” sign tells us that the largest factor must be negative

( 6)( 1)x x

To determine the number factors we must find the factors of 6 that subtract to give 5.

Those factors are 6 and 1.

Example #3

2Factor 2 15x x

How would we factor a quadratic trinomial whose leading coefficient is not 1? Consider the following problem.

To begin, set up 2 sets of parenthesis placing the leading coefficient 2 and the variable x in the first position in each set of parenthesis.

Note: We have an extra 2 as a factor because 2X2=4!

(2 )(2 )x x

Example #3 (continued)

22 15x x Now, apply the rule of signs that was mentioned previously. The signs must be different and the largest factor must be positive.

Next, multiply the constant (15) and the leading coefficient (2) to get 30. We now need the factors of 30 that subtract to equal 1 (middle coefficient)

6X5 = 30 and 6 – 5 = 1

(2 )(2 )x x

Example #3 (continued)

22 15x x

(2 )(2 )x x (2 6)(2 5)x x

Now insert the factors of 6 and 5 below

We must now remove the extra factor of 2 that was inserted at the beginning of the problem. Notice that 2 is a GCF in the first set of parenthesis,

2(x +3)(2x−5)Cross out the 2 and what remains is the answer!

Checking Example #3

22 15x x 2First 2x

22 15x x

Now, let’s check the previous example by using the FOIL Method.

Combining similar terms we get

Outer 5x Inner 6x Last 15

( 3)(2 5)x x

It checks!!

Let’s Try Another!Example #4

23 10 8x x (3 - )(3 )x x

(3 6)(3 4)x x 3( 2)(3 4)x x

( 2)(3 4)x x

Place the 3x as the first term in each set of parenthesis.

Find the factors of 24 (8X3) that ADD up to 10.

Factor out the extra 3 and cross it out.

Answer !!

The signs must be the same and both negative.!

One Last Example

22 5 7x x (2 )(2 )x x We now try to find the

factors of 14 (2X7) that ADD up to 5!

There are none !!!

So, the Quadratic Trinomial is PRIME and cannot be factored.

This method of factoring will be used to solve quadratic equations and will also help in graphing quadratic functions and inequalities.