Zero product property

What is the first step in solving

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

• Plug in zero for x

• Solve for x

• distribute

• Set each binomial factor equal to zero

Tags

CCSS.HSA-REI.B.4B

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

Solve and find your x-intercepts:
0=x(x - 6)

• x=0 and x=-6
• x=1 and x=-6
• x=0 and x=6
• x=6

Tags

CCSS.HSA-REI.B.4B

Factor: x² – 5x – 24

• (x - 5)(x + 19)
• (x - 8)(x + 3)
• (x - 3)(x + 8)
• (x - 19)(x + 5)

Tags

CCSS.HSA.REI.B.4

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor

4x² - 25

• (2x + 5) (2x - 5)

• (2x - 5)²

• (2x + 5)²

• 2x + 5(2x - 5)

Tags

CCSS.HSA.APR.C.4

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

• x = -5 and 4

• x = -5 and -4

• x = 5 and -4

• x = 5 and -4

Tags

CCSS.HSA-REI.B.4B

Factor
n² + 16n + 63

• (n-7)(n+4)
• (n+7)(n-9)
• (n-3)(n-10)
• (n+7)(n+9)

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

What are the values of x when (2x+5)(x-6)=0?

• -6, 5/2

• -5/2, 6

• 6, 2

• 5, -6

Tags

CCSS.HSA-REI.B.4B

 $j\left(j-9\right)=0$  

•  $j=0,\ j=9$  

•  $j=0,\ j=-9$  

•  $j=9$  

•  $j=-9$  

Tags

CCSS.HSA-REI.B.4B

 $\left(c+6\right)\left(c-7\right)=0$  

•  $c=-6,\ c=7$  

•  $c=6,\ c=-7$  

•  $c=6,\ c=7$  

•  $c=-6,\ c=-7$  

Tags

CCSS.HSA-REI.B.4B