Question

Which shows one way to determine the factors of x3 – 12x2 – 2x + 24 by grouping?

x(x2 – 12) + 2(x2 – 12)
x(x2 – 12) – 2(x2 – 12)
x2(x – 12) + 2(x – 12)
x2(x – 12) – 2(x – 12)




Which shows one way to determine the factors of 12x3 – 2x2 + 18x – 3 by grouping?

2x2(6x – 1) + 3(6x – 1)
2x2(6x – 1) – 3(6x – 1)
6x(2x2 – 3) – 1(2x2 – 3)
6x(2x2 + 3) + 1(2x2 + 3)



Factor x3 + x2 + x + 1 by grouping. What is the resulting expression?

(x2)(x + 1)
(x2 + 1)(x)
(x2 + 1)(x + 1)
(x3 + 1)(x + 1)



The polynomial 10x3 + 35x2 − 4x − 14 is factored by grouping.

10x3 + 35x2 − 4x − 14

5x2(____) − 2(____)

What is the common factor that is missing from both sets of parentheses?

−2x − 7
2x + 7
−2x2 + 7
2x2 + 7



Faelyn grouped the terms and factored the GCF out of the groups of the polynomial 6x4 – 8x2 + 3x2 + 4. Her work is shown.

Step 1: (6x4 – 8x2) + (3x2 + 4)

Step 2: 2x2(3x2 – 4) + 1(3x2 + 4)

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

Faelyn should realize that her work shows that the polynomial is prime.
Faelyn should go back and regroup the terms in Step 1 as (6x4 + 3x2) – (8x2 + 4).
In Step 2, Faelyn should factor only 2x out of the first expression.
Faelyn should factor out a negative from one of the groups so the binomials will be the same.

Answer 1

1) x^2(x − 12) − 2(x − 12).
(Group as (x^3 − 12x^2) + (−2x + 24).)

2) 2x^2(6x − 1) + 3(6x − 1).
(Group as (12x^3 − 2x^2) + (18x − 3).)

3) (x^2 + 1)(x + 1).
(x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1).)

4) 2x + 7.
(10x^3 + 35x^2 − 4x − 14 = 5x^2(2x+7) − 2(2x+7).)

5) Faelyn should go back and regroup the terms as (6x^4 + 3x^2) − (8x^2 + 4).
Then 3x^2(2x^2 + 1) − 4(2x^2 + 1) = (3x^2 − 4)(2x^2 + 1).