Factoring and Solving Quadratics

Factoring and Solving Quadratics

We are going to continue reviewing factoring and extend that idea into SOLVING! :)

Multiple Choice

In order to factor the trinomial:  

 $y=x^2+bx+c$   
we need to find to integers that....?

• multiply to b

• multiply to c

• divide by b

• divide by c

Multiple Choice

In order to factor the trinomial:  

 $y=x^2+bx+c$   
we need to find to integers that....?

• add to b

• add to c

• subtract to b

• subtract to c

As an example, if we have x² + 7x + 12

Integers that sum to 7 are:   0+7, 1+6, 2+5, 3+4

Factors of 12 are: 1⋅12, 2⋅6, 3⋅4

  

The only pair that are in common are 3 and 4.  

This means we can factor it to:   (x+3)(x+4)

  

Multiple Choice

Factor:

 $x^2+9x\ -36$  

• (x + 12)(x  - 3)

• (x + 4)(x + 9)

• (x - 12)(x  - 3)

• (x - 12)(x  + 3)

Multiple Choice

Factor:

 $x^2+5x\ -6$  

• (x + 5)(x  - 6)

• (x + 2)(x + 3)

• (x - 6)(x  - 1)

• (x + 6)(x  - 1)

Multiple Choice

Factor:

 $x^2+7x\ -30$  

• (x + 5)(x  - 6)

• (x + 10)(x + 3)

• (x - 3)(x + 10)

• (x  - 10)(x  - 3)

To solve:

Take each pair of parenthesis.

Create an equation with it.

Solve that equation.

(x-4) =0 and (x+5)=0

What would the answers be?

Multiple Choice

Solve for the values of x:

(x + 2)(x - 3) = 0

• x = 2, 3

• x = -2, -3

• x = 2, 3

• x = -2, 3

Multiple Choice

Solve for the values of x:

(x - 5)(x - 1) = 0

• x = 5, 1

• x = 5, -1

• x = -5, 1

• x = -5, -1

Multiple Choice

Solve by factoring: x² + 3x - 18 = 0

• x = -3 and x = 6

• x = 3 and x = -6

• x = -9 and x = 2

• x = 9 and x = -2