can you help me with a math problem?

Question

Factor completely and select the correct answer.

3x^5 + 15x^3 – 108x

(x – 2)(x^2 – 1) – 6x – 6

Answer 1

The problems ask for you to factor the expressions completely...

PROBLEM 1:

Factor out a common 3x first:
3x(x^4 - 5x^2 - 36)

Now I would notice that if you replace k = x², then you can factor the item in parentheses as:

3x(k - 9)(k + 4)
3x(x² - 9)(x² + 4)

Finally, you have a difference of squares in the first parentheses so use that rule a² - b² = (a - b)(a + b)
3x(x - 3)(x + 3)(x² + 4)

PROBLEM 2:

Expand the difference of squares:
(x - 2)(x² - 1) - 6x - 6

Factor out a -6 from the last two terms:
(x - 2)(x - 1)(x + 1) - 6(x + 1)

Factor out a common x + 1:
[(x - 2)(x - 1) - 6 ] (x + 1)

Now expand the expresion in the brackets:
[(x² - 3x + 2) - 6 ] (x + 1)
( x² - 3x - 4 ) (x + 1)

Now you can factor the first expression:
[ (x - 4)(x + 1) ](x + 1)

Then simplify to:
(x - 4)(x + 1)²

Answer 2

3x(x^4 + 5x^2 - 36)


x^3 - x - 2x^2 + 2 - 6x - 6
x^3 - 2x^2 - 7x - 4

Answer 3

dnt get it ... u want them in an equation = 0
or wt?

Answer 4

3x^5+15x^3-108x
3x(x^4+5x^2-36)
3x(x^2+9)(x^2-4)
3x(x^2+9)(x+2)(x-2)

(x-2)(x^2-1)-6x-6
(x-2)(x+1)(x-1)-6(x+1)
(x+1){(x-2)(x-1)-6)}
(x+1){x^2-3x+2-6}
(x+1){(x^2-3x-4)}
(x+1)(x-4)(x+1)
(x+1)^2(x-4)