Factoring by Grouping

Omar grouped the terms and factored the GCF out of the groups of the polynomial 3x³ – 15x² – 4x + 20. His work is shown.

Step 1: (3x³ – 15x²) + (–4x + 20)

Step 2: 3x²(x – 5) + 4(–x + 5)

Omar noticed that he does not have a common factor. Which accurately describes what Omar should do next?

• Omar should realize that his work shows that the polynomial is prime.

• Omar should go back and regroup the terms in Step 1 as (3x³ – 15x²) – (4x + 20).

• In Step 2, Omar should factor only out of the first expression.

• Omar should factor out a negative from one of the groups so the binomials will be the same.

20x³-25x²+4x-5

• (5x²-5)(4x+1)
• 5(x²-1)(4x+5)
• (5x²+1)(4x-5)
• (5x²+1)(4x-1)

2x³-8x²-x+4

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

x³-5x²+5x-25

• (x²-5)(x-5)
• (x²+5)(x-5)
• (x²+5)(x+5)
• (x²-1)(x-25)

What is the factored form of x³+7x²-2x-14

• (x²-2)(x-7)
• (x²+2)(x-7)
• (x²+2)(x+7)
• (x²-2)(x+7)

Factor by grouping:
4p³+8p²+3p+6

• (2p²+3)(p+2)
• (2p²+6)(p+4)
• (4p²+3)(p+2)
• (4p²+8p)(3p+6)

What method should you use to factor the polynomial?

5n³-10n²+3n-6

• Factor by Grouping

• Factor out the GCF

• Difference of Squares

• Master Product

What is the factored form of 5n³-10n²+3n-6

• (5n²+3)(n-2)

• (5n²-3)(n-2)

• (5n³+3)(n-2)

• 10n(n²-6)

Factor by grouping:
3x³-x²+6x-2

• (x²+2)(3x+1)
•  (x²+2)(3x-1)
• (3x³+6x)(x²-2)
• none of the aboe

Factor completely:

5n³-10n²+3n-6

• (5n²+3)(n-2)

• (5n²-3)(n-2)

• (5n³+3)(n-2)

• 10n(n²-6)