Algebra 2: Solving Quadratics by Factoring

Solve:  $y=\left(x+1\right)\left(x-5\right)$  

• x = 1 and x = - 5

• x= - 1 and x = 5

• x = 1 and x = 5

• x = -1 and x = - 5

Solve:  $y=\left(x-3\right)\left(x+3\right)$  

• x = 3

• x = - 3

• x = 3 and x = - 3

• x = 9

Solve:  $y=\left(2x-3\right)\left(x-4\right)$  

• x =  $\frac{2}{3}$  and x = 4

• x =  $-\frac{2}{3}$  and x = - 4

• x =  $\frac{3}{2}$  and x = 4

• x =  $-\frac{3}{2}$  and x = - 4

Solve by factoring:  $y=x^2+9x+8$  

• x = 8 and x = 1

• x = 4 and x = 2

• x = - 8 and x = -1

• x = - 4 and x = - 2

Solve by factoring:  $y=x^2-3x-18$  

• x = 6 and x = - 3

• x = - 6 and x = 3

• x = 9 and x = 2

• x = - 9 and x = 2 

Solve by factoring:  $y=x^2-12x+36$  

• x = 9 and x = - 4

• x = 6 

• x = - 6

• x = 6 and x = - 6

Solve by Factoring:  $y=x^2+13x+42$  

• x = 21 and x = 2

• x = 7 and x = 6

• x = - 7 and x = - 6

• x = - 14 and x = 3

Solve by factoring:  $y=x^2-81$  

• x = 9

• x = - 9 

• x = 9 and x = - 9

• x = 27 and x = 3

What are the factors AND solutions of x² + 2x – 3 = 0

• (x - 2)(x + 1); x=2, x=-1
• (x + 1)(x - 3); x=-1, x=3
• (x + 2)(x - 1); x=-2, x=1
• (x - 1)(x + 3); x=1, x=-3

Tags

CCSS.HSA.APR.B.3

CCSS.HSA.REI.B.4

What is the first step in solving 
(x + 2)(x - 3) = 0      ?

• Plug in zero for x
• Solve for x
• FOIL
• Set each binomial factor equal to zero

Tags

CCSS.HSA-REI.B.4B