BIM Alg1 7.6: AC Factoring of Quadratic Equations

BIM Alg1 7.6: AC Factoring of Quadratic Equations

AC Method of Factoring when you have an a value on equation

 $ax^2+bx+c$  

Step 1:
Identify the a and c.

 $4x^2+13x+3$  
a=4 and c=3

Step 2:

Multiply a*c

Ex: a*c=4*3=12

Step 3:

Identify factors of ac that when added give you b.

Ex: 4*3=12 b=13

12+1=13

6+2=8

4*3=7

Step 4:

 Set-up New Eq:  

 $4x^2\ \ \ \ \ \ \ \ +3$  

Then we put out factors in
 $4x^2+1x+12x+3$  

Step 5: We split out equation.


 $\left(4x^2+1x\right)+\left(12x++3\right)$  

Step 6: 

We pull out or undistribute common factors.

 $x\left(4x+1\right)+3\left(4x+1\right)$  

Step 7: 

Notice we have a common factor in both. Mike starts talking about meat and potatoes and that he is hungry???

 $\left(x+3\right)\left(4x+1\right)\ are\ our\ factors!$  

Step 8: Check our Solution

(x+3)(4x+1)= 

 $4x^2+x+12x+3$  
 $4x^2+13x+3\ \ \ Check$  

What if we had swapped the factors when we remade the equation?

 $\left(4x2+12x\right)+\left(1x+3\right)$  
4x((x+3)+1(x+3)
(4x+1)(x+3)  Eureka or 
"Meat and Potatoes!"

Ex 2: WITH A (-b).
 $3x^2-7+2$  
Solve on Desktops

 $3x^2-7x+2$  
Step 1:  a=3 and c=2
Step 2:  ac=3*2=6
Step 3:  factors of 6 that add to b.  (b=-7)
-6 and -1
Step 4:  New equation
 $3x^2-6x-1x+2$  

Step 5: Divide equation

 $\left(3x^2-6x\right)-1\left(x-2\right)$  

Step 6: Pull out common factors

3x(x-2)-1(x-2)

We shout "Meat and Potatoes!"

Step 7: We simplify

(3x-1)(x-2)

Step 8: We check out work
(3x-1)(x-2)=

 $3x^2-6x-x+2$   Simplify

 $3x^2-7x+2$  Check

Ex3: Negative ac

 $2x^2-5x-7$ 
What is our ac? 

a=2 and c=-7

ac=-14

What factors equal b?

b=-5

factors of -14 that equal -5?

-7 and +2

What will our new equation look like?

 $2x^2-7x+2x-7$  

 $\left(2x^2-7x\right)+\left(2x-7\right)$  
x(2x-7)+1(2x-7) M&P!
(x+1)(2x-7)
Check your answer

 $-4x^2-8x+5$  

What happens when a is negative?

What is a and c?

What is the ac value?

What are the factors that =b?

What is our new equation?

 $-4x^2+2x-10x+5\$  or

 $-4x^2-10x+2x+5$  

What solution did you get?

Work it out on the table!

I want to hear Meat and Potatoes when you get your new solution!

 $-4x^2+2x-10x+5$  
2x(-2x+1)+5(-2x+1) M&P!
(2x+5)(-2x+1)

 $-4x^2-10x+2x+5$  
2x(-2x-5)-(-2x+5) M&P!
(2x-1)(-2x-5)

What...the factors are different?

(2x+5)(-2x+1) vs (2x-1)(-2x-5)

If you set them =0, same solution

x=5/2, x=1/2

Interesting!

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