Factoring completely...i need help please?

Question

1. Factor completely: 16x^4 - 81

A. (2x - 3)^2(4x^2 + 9)

B. (4x^2 + 9)(4x^2 - 9)

C. (2x + 3)(2x + 3)(4x^2 - 9)

D. (2x + 3)(2x - 3)(4x^2 + 9)






2. Factor by grouping: 8x^3 - 12x^2 + 6x - 9.


3. Factor completely - 49 + x^2.

A. ( -7 + x)( -7 - x)

B. ( -7 + x)(7 + x)

C. (x - 7)(x - 7)

D. prime


4. Factor completely 5x^3 - 5x^2 - x + 1.

A. 5x^2(x - 1) - 1(x - 1)

B. 5x^2(x - 1)^2

C. (x - 1)(5x^2- 1)

D. prime


5. Factor completely 4x^2 + 9.

A. (2x + 3)(2x + 3)

B. (2x - 3)(2x - 3)

C. (2x + 3)(2x - 3)

D. prime

Answer 1

1. D. (2x + 3)(2x - 3)(4x^2 + 9)
3. A. ( -7 + x)( -7 - x)
5. D

Answer 2

1. This equation is a perfect square.
(4x^2 + 9) (4x^2 - 9 )
However, there is another perfect square in the second parenthesis, due to the subtraction sign. So the answer is:
(4x^2 + 9) (2x + 3) (2x - 3) D

2.
(8x^3 - 12x^2) (6x - 9)
4x^2(2x - 3) 3(2x - 3)
So the answer is:
(4x^2 + 3) (2x - 3)

3.This equation is a perfect square. The answer is:
(-7 + x) (7 + x) B

4. Factor this equation by grouping.
(5x^3 - 5x^2) (-x +1)
5x^2(x - 1) -1(x - 1)
So the answer is:
(5x^2 -1) (x - 1) C.

5.This equation would be a perfect square if subtraction was involved, but this is an addition problem that cannot factor. The answer to this problem is:
PRIME D.

I hope I helped!!

Answer 3

Factoring completely...i need help please?
1. Factor completely: 16x^4 - 81

A. (2x - 3)^2(4x^2 + 9)

B. (4x^2 + 9)(4x^2 - 9)

C. (2x + 3)(2x + 3)(4x^2 - 9)

D. (2x + 3)(2x - 3)(4x^2 + 9)


16x^4 - 81 = 4x^2 + 9 multiplied by 4x^2 - 9

4x^2 + 9 is not factorable but 4x^2 - 9 can be written as 2x + 3 and 2x - 3

So, D is the right answer.

Do the other problems on similar lines.

The last question is answered above. It is not a factorable expression, may be a prime.