Solving Polynomials That Are Factored Already

Solve for the values of x:
(2x + 5)(5x - 2) = 0

• { -5/2, -2/5 }
• { -5/2, 2/5 }
• { 5/2, 2/5 }
• { 5/2, -2/5 }

Tags

CCSS.HSA-REI.B.4B

Solve for the values of x:
(2x - 1)(x + 4) = 0

• { 1/2, -4 }
•  { 1/2, 4 }
• { 2, -4 }
• { -2, 4 }

Tags

CCSS.HSA-REI.B.4B

Solve for the values of x:
(x - 5)(x - 1) = 0

• { 5, 1 }
• { 5, -1 }
• { -5, 1 }
• { -5, -1 }

Tags

CCSS.HSA-REI.B.4B

Solve for the values of x:
(x - 3)(x + 6) = 0

• { -3, -6 }
• { 3, -6 }
• { -3, -6 }
• { 3, 6 }

Tags

CCSS.HSA-REI.B.4B

What is the name of this property?

• Zero Exponent Property
• Zero Product Property
• Commutative Property
• Zero Sum Property

Tags

CCSS.8.EE.C.7A

Solve for the values of x:
(x + 2)(x - 3) = 0

• { 2, 3 }
• { -2, -3 }
• { 2, 3 }
• { -2, 3 }

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 (2x -8) (x + 5) = 0 

• x = 5, x = 8
• x= -4, x = 5
• x= 4, x = -5
• x= -5, x = -8

Tags

CCSS.HSA-REI.B.4B

Tags

CCSS.HSA-REI.B.4B

Tags

CCSS.HSA-REI.B.4B