Factoring and Solving Polynomials; Dividing Polynomials

x² + 12x + 32 = 0

• x = -4, 8

• x = 4, 8

• x = -4, -8

• x = 0, 12

Tags

CCSS.HSA-REI.B.4B

m²-10m=0

• m=0 and m=-10
• m=1 and m=10
• m=0 and m=10
• m=1 and m=-10

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

What is the first step when factoring a polynomial?

• How many terms does it have?

• Is there a GCF?

• What strategy should I use?

• Why do I have to learn this?!?!?!

Answer explanation

Knowing how many terms you have, figuring out what strategy to use, and seeing if there is a GCF are equally important.

Tags

CCSS.HSA.SSE.A.2

Factor completely:
4x²y - 12xy + 5y

• (2xy - 1) (2x - 5)
• prime
• y (2x - 1) (2x - 5)
• 4 (xy - 5) (x - 1)

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

CCSS.HSA.APR.A.1

Factor completely:
4x² - 121

• (2x + 11) (11 - 2x)
• (2x - 11) (2x - 11)
• (2x + 11) (2x + 11)
• (2x + 11) (2x - 11)

Tags

CCSS.HSA.APR.C.4

Factor
4n²-17n-15

• (n-10)(6n+1)
• (4n-5)(n-3)
• (n-5)(4n-3)
• (n-5)(4n+3)

Tags

CCSS.HSA.APR.B.2

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.  
8x³ - 18x

• 2x(4x²-9)                                    
• (4x²+3)(2x-3)
• 2x(2x+3)(2x-3)
• PRIME

Tags

CCSS.HSA.APR.C.4

x² + 81

• Prime
• ( x - 9 ) ( x + 9 ) 
• ( x + 9 ) ( x + 9 ) 
• ( x - 9 ) ( x - 9 ) 

Tags

CCSS.HSA.APR.C.4

Factor Completely:

2n³ + 7n² - 2n - 7

• cannot be factored

• (n - 1)²(2n + 7)

• (n² - 1)(2n + 7)

• (n + 1)(n - 1)(2n + 7)

Tags

CCSS.HSA.APR.B.2

CCSS.HSA.APR.B.3

CCSS.HSA.SSE.A.2