Algebra 2 Urgent: Polynomial functions?

Question

how would this work because of the "i"?

-2, -2, 3, -4i

Answer 1

These are the roots to the following quartic polynomial:

(x + 2)(x + 2)(x - 3)(x + 4i)

= (x^2 + 4x + 4)(x^2 + 4ix - 3x - 12i)

= x^4 + (4i + 1)x^3 + (4i - 8)x^2 + (32i)x^2 - 12x - 48i

where i = √-1

This polynomial looks a bit weird as the roots do not have conjugate pairs. So if the conjugate pair, +4i, is also a root, then it will be a quintic polynomial, with no imaginary coefficients:

(x + 2)(x + 2)(x - 3)(x + 4i)(x - 4i)

= (x^2 + 2x + 1)(x^3 + 16x - 3x^2 - 48)

= x^5 - x^4 + 11x^3 - 19x^2 + 0x - 48

Answer 2

Hmmm.. I'm not sure what this question is asking, so I'll try and read your mind.

If you want a polynomial of degree 4 that contains exactly these roots, just expand

p(x) = (x + 2)^2 (x - 3)(x + 4i)

However, if you want the smallest degree polynomial with real coefficients, then, since complex roots come in pairs, 4i is also a root.

p(x) = (x + 2)^2 (x - 3)(x + 4i)(x - 4i)

p(x) = (x^2 + 2x + 1)(x - 3)(x^2 + 16)

p(x) = (x^2 + 2x + 1)(x^3 + 16x - 3x^2 - 48)

p(x) = x^5 + 16x^3 - 3x^4 - 48x^2 + 2x^4 + 32x^2 - 6x^3 - 96x
+ x^3 + 16x - 3x^2 - 48

p(x) = x^5 - x^4 + 11x^3 - 19x^2 - 48

Answer 3

There has to be a fifth zero because imaginary solutions travel in pairs therefore if you have -4i +4i has to be a zero also.

f(x) = (x + 2)(x + 2)(x - 3)(x + 4i)(x - 4i)

Answer 4

math fr... is correct as long as you want an imaginery polynomial function. There is a therom that for real terms, the imaginery root have to be in pairs. So if that is what you are looking for, these roots won't get you there.

Answer 5

K the "i" represents the square root of -1. So itd be 4 I think.

Answer 6

Could you state your question more specifically?

Answer 7

i means sqrt(-1)