i am really stuck on one problem and i dont evne know how to aproach it:
Write a polynomial function f of least degree that has rationa coefficients, a leading coefficient of 1, and the zeros 2, -3 and -3I
so its
(x-2)(x+3)(x+3i) now thats how far i got cuz i have NOOOOO idea how to get rid of that darn i
2007-01-16 05:01:17 · 7 answers · asked by forza_milan_campione 1 in Science & Mathematics ➔ Mathematics
i think the first person got it right on target. i just needed to add that but i though it would mess up the whole equation. thanks i got it now.
2007-01-16 05:25:38 · update #1
You're close. All you need to do is put in another factor of (x-3i) so that you'll have a product of conjugates which is a difference of squares. In the case of complex conjugates it comes out looking like a -sum- of squares.
This is not cheating, to put in that extra factor. Complex roots of real coeffiecnt polynomials ALWAYS come in conjugate pairs. That is, you -can't- have a -3i without a 3i.
Try it and see.
(x-2)(x+3)(x-3i)(x+3i)
2007-01-16 05:05:10 · answer #1 · answered by Joni DaNerd 6 · 1⤊ 0⤋
i'm at a loss for words. Do the teaspoon factors should be measured with the cup? If no longer, it truly is the a million/4 cup because all of the different measurements are divisible by a million/4. That appears like a peculiar question besides the undeniable fact that.
2016-11-24 21:22:24 · answer #2 · answered by schifano 4 · 0⤊ 0⤋
If a polynomial with rational coef. has the complex root a+bi it also must have the cojugate a -bi.So your polinom must be at least
(x-2)(x+3)(x+3i)(x-3i) = (x-2)(x+3)(x^2+9) which complies
2007-01-16 05:57:44 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
first, multiply (x-2)(x+3)
you have (x^2 +x - 6).
then, multiply (x^2 +x - 6)(x+3i)
Now the problem is you can not multiply those two together, so you gonna have to ask your teacher if the question is correct. If your says the question is correct, and your teacher gave you the correct answers, please i do like to know what the answer is.
I think the question is wrong.
2007-01-16 05:15:08 · answer #4 · answered by bochevik 2 · 0⤊ 0⤋
You need the conjugate (x-3i) because, when the polynomial has real coefficients, the complex roots come in pairs, (a+bi) and (a-bi).
2007-01-16 05:06:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋
multiply them. I got the answer as
(x)^3 + 3(x)^2i+X^2+3xi-6x-18i
2007-01-16 05:07:45 · answer #6 · answered by ♥♪♫Priya_akki™♫♪♥ 6 · 0⤊ 0⤋
distribute it or somthing
2007-01-16 05:05:18 · answer #7 · answered by monkeytits 2 · 0⤊ 2⤋