Factor Completely

Factor Completely:
3x⁵ - 27x³

• 3(x³ - 3)(x² + 3)
• prime
• 3x³(x² - 9)
• 3x³(x - 3)(x + 3)

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor completely 
 6m² + 16m

• 2(3m² + 8m)
• 2m(3m + 8)
• 2m(3m - 8m)
• not factorable

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.
 $4x^2+24x-64$  

•  $4\left(x+8\right)\left(x-2\right)$  

•  $4(x^2+6x-16)$  

•  $4\left(x^2+24x-16\right)$  

•  $4\left(x-8\right)\left(x+2\right)$  

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.
 $25x^2-100$  

•  $5\left(x+4\right)\left(x-4\right)$  

•  $5\left(x^2-20\right)$  

•  $25\left(x^2-4\right)$  

•  $25\left(x+2\right)\left(x-2\right)$  

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.
 $8x^2-200$  

•  $8\left(x+5\right)\left(x-5\right)$  

•  $8\left(x^2-25\right)$  

•  $4\left(2x^2-50\right)$  

•  $4\left(2x+25\right)\left(2x-25\right)$  

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.
 $5x^3-20x$  

•  $x\left(5x^2-20\right)$  

•  $5x\left(x^2-4\right)$  

•  $5x\left(x+2\right)\left(x-2\right)$  

•  $5x\left(x+4\right)\left(x-4\right)$  

Tags

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely.
 $4x^2-24x+36$  

•  $4\left(x+3\right)\left(x-3\right)$  

•  $4\left(x^2-6x+9\right)$  

•  $4\left(x-3\right)^2$  

•  $4\left(x+3\right)^2$  

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Find the expression equivalent to 8x⁴+22x³-6x²

• 2x²(4x-1)(x+3)
• 2x²(4x-1)(x-3)
• 2x²(4x+1)(x-3)
• 2x(4x-1)(x+3)

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

Factor Completely: 
x⁴-9x²

• x²(x+3)(x-3)
• (x²+3x)(x²-3x)
• (x²+9)(x-3)(x+3)
• x²(x²-9)

Tags

CCSS.HSA.SSE.A.2

What went wrong? (Select all errors that occured)

• Step 1: GCF not factored out properly $2\left(x^{2\ }-x-24\right)$  

• Step 2: Binomials not factored properly  $2\left(2x-12\right)\left(2x+8\right)$  

• Step 1: GCF not factored out properly  $2\left(x^2-x-12\right)$  

• Step 2: Binomials not factored properly   $2\left(x-4\right)\left(x+3\right)$