how do u factor these..?

Question

2x^3+8x^2+8x

2x^4-32

x^2+16

disregard the second on i already solved it

Answer 1

2x^3 + 8x^2 + 8x

=2x(x^2 + 4x + 4)

=2x(x + 2)^2

2x^4 - 32

=2(x^4 - 16)

=2(x^2 + 16)(x^2 - 16)

=2(x^2 + 16)(x + 4)(x - 4)

x^2 + 16 has no rational factors.

Hope this helps, Twiggy.

Answer 2

well, for the first one you can factor out 2x, so you have:

2x(x^2 + 4x +4), then you can factor the polynomial by finding two numbers that multiply out to 4 and add up to 4, obviously 2 and 2 so we have:

2x(x+2)(x+2) or 2x(x+2)^2 (since we have two its squared)

the third one is prime, but it could be factored on the complex plane, the solutions will be complex numbers:

x^2 + 16 = 0
x^2 = -16

now take the square root of both sides

x = +/-4i (where i is the imaginary unit: the square root of -1, note also that when you take the square root of a number, you always get a negative and a positive)

Answer 3

2x^3 + 8x^2 + 8x

2x (x^2 + 4x + 4)

2x (x + 2) (x + 2)


2x^4 - 32

2 (x^4 -16)

2 (x^2 + 4) (x^2 - 4)

2 (x^2 + 4) (x - 2) (x + 2)

x^2 + 16 (I think this one is non-factorable)

Answer 4

2x^3 + 3x^2 + 8x
x(2x^2 + 8x +8)
x(2x + 4)(x + 2)

2x^4 - 32
(2x^2 - 8)(x^2 + 4)