SOLUTION: I NEED HELP ASAP Can you break this problem down for me so, I can do other problems like it Find a polynomial function of lowest degree with rational coefficients that has th

Question 238136: I NEED HELP ASAP
Can you break this problem down for me so, I can do other problems like it
Find a polynomial function of lowest degree with rational coefficients that has the given numbers of some of its zeros.
2+i, 3
I know there is a conjugate of 2+i, which is 2-i. Which makes the degree of the polynomial 3 and the number of zeros 3
2+1,2-i,3
(x-(2+i))(x-(2-i))(x-3)
I can't get passed the imaginary numbers' portion of the problem. And if I do, it is wrong.
PLEASE HELP

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
two roots of this equation are:

x = 3
x = 2+i

x - 3 = 0
or

x - 2 - i = 0

it seems you're on the right track with the conjugate of x = 2-i

that should make them cancel out when you multiply them together.

let's do that and see how it works.

x = 2-i means that x - 2 + i = 0

you have 3 factors.

they are:

x - 3 = 0
x - 2 + i = 0
x - 2 - i = 0

first you want to multiply x - 2 + i by x - 2 - i

each term in one factor has to be multiplied by each term in the other factor.

start with x in the left factor and multiply each term in the right factor from left to right.

x * x = x^2
x * -2 = -2x
x * -i = -ix

do -2 next.

-2 * x = -2x
-2 * -2 = 4
-2 * -i = 2i

do i next.

i * x = ix
i * -2 = -2i
i * -i = -i^2 = 1 because i^2 = -1

add all your terms together to get:

x^2 - 2x - ix -2x + 4 + 2i + ix -2i + 1

+ ix and - ix cancel out
+ 2i and - 2i cancel out
-2x and -2x become -4x
+ 4 and + 1 become + 5

your equation becomes x^2 - 4x + 5

multiply this equation by x-3 to get:

x * x^2 = x^3
x * -4x = -4x^2
x * 5 = 5x
-3 * x^2 = -3x^2
-3 * -4x = 12x
-3 * 5 = -15

add these terms together to get:

x^3 - 4x^2 + 5x - 3x^2 + 12x - 15

combine like terms to get:

x^3 -7x^2 + 17x - 15

that's your equation.

a graph of this equation looks like:

It only has one real root (x-axis crossing point).

If you set x = 3, the equation should become equal to 0.

I confirmed that it does.

If you set x = 2-i, the equation should become equal to 0.

I confirmed that it does. It was a nosebleed but it was confirmed.

I suspect 2+i will confirm as well although I'm not going to go through that again.

Not today, anyway.

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