Question

Does anyone know how to solve complex zeros.?

Answer 1

hit meh!

Answer 2

Use this website :) http://coolmath.com/algebra/22-graphing-polynomials/07-complex-zeros-02.htm its kiddy but it will help :)

Answer 3

2x^3-3x^2+8x-12=0 @Brostep0s

Answer 4

And okay, ill try it. Thanks. (: @Notamathgenius

Answer 5

You're most certainly welcome :)

Answer 6

is it a function?

Answer 7

@aha.its.kristen

Answer 8

@Brostep0s I'm not sure. The question is

Solve 2x^3-3x^2+8x-12=0

Answer 9

do you know synthetic division?

Answer 10

(2x^3 - 3x^2 - 8x + 12) / (x - 2) = 0

First, let's observe that since (x - 2) is in the denominator, x = 2 is not in our domain, and therefore cannot be a solution even if algebraically it comes up as one. From here, the first step is to look at just the numerator:

2x^3 - 3x^2 - 8x + 12

We need to factor this to find solutions, so first, group (remembering to factor out the negative sign):

(2x^3 - 3x^2) - (8x - 12)

Factor each group:

x^2 * (2x - 3) - 4 * (2x - 3)

Since there is a common term, we can recombine:

(x^2 - 4) * (2x - 3)

Now, this first term is the difference of two squares and so it can further be factored to get the final factored form:

(x + 2) * (x - 2) * (2x - 3)

We can now put this back over our denominator:

((x + 2) * (x - 2) * (2x - 3)) / (x - 2) = 0

Since we have an (x - 2) term on the top and bottom, they cancel, leaving:

(x + 2) * (2x - 3) = 0

From here, we set each term equal to 0 independently, and we solve:

x + 2 = 0 and 2x - 3 = 0

Solving each, we get two solutions for the zeros:

x = -2 and x = 3/2

Answer 11

Oh my god, thank you so much. (: @Brostep0s

Answer 12

yeppers cx

Answer 13

did you fan me? jw cx

Answer 14

Yes i did (: @Brostep0s

Answer 15

I seen that... xD

Answer 16

Yeah, message me? (:

Answer 17

hmu cx