Alg. 2 Ch. 4 Test (Final) Factoring Quadratics/Transformations

Solve for X:

(x-5)(x+6)=0

• {5,5 }

• { -5,-5 }

• { -5,-6 }

• {5,-6 }

Tags

CCSS.HSA-REI.B.4B

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

Which equation matches these solutions? x=5,-7

• (x+5)(x-7)=0
• (x-5)(x+7)=0
• (x+5)(x+7)=0
• (x-5)(x-7)=0

Tags

CCSS.HSA-REI.B.4B

Solve the quadratic equation.

• -2, 4

• -8

• ±8

• No real solution

Tags

CCSS.HSA-REI.B.4B

Use the quadratic formula to solve 2x² + 2x - 12 = 0.

• -2, 3

• 2, 3

• 2, -3

• -2, -3

Tags

CCSS.HSA-REI.B.4B

Solve the quadratic equation.

• -4, 2

• -2, 4

• -16

• -4

Tags

CCSS.HSA-REI.B.4B

Convert x²+6x+9=0 to factored form and identify the solution.

• (x+3)(x+3)=0; x=-3

• (x+4)(x+5)=0; x=-4, -5

• (x-3)(x-3)=0; x=3

• Not factorable

Tags

CCSS.HSA-REI.B.4B

Find the zeros of this quadratic. (Finding the zeros means to find the solutions.)

• b=4/5, b=3

• b=4, b=-3

• b=1/5, b=3

• b=4/5, b=-3

Tags

CCSS.HSA-REI.B.4B

Solve by factoring:
x² - 9x = 0

• 9
• 0, 9
• -9, 0
• -9

Tags

CCSS.HSA-REI.B.4B

Factors AND find the 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-REI.B.4B