can you help me find the equation of this?

Question

we are asked to find the equation given only its roots..

these are the roots:
x = 3, ( 1 + i), ( 1 – i)

I just can't solve becaise of its imaginary sign..
please help..
thanks!

Answer 1

Any time you get something like this, just write the factors as (x-r1)(x-r2)(x-r3) however many roots you have.
Don't let the complex numbers scare you, notice that they come in conjugate pairs. That is, same numbers, just different + / - signs. Any time you multiply together two conjugates you get the difference of two squares. (that is, (x+y)(x-y) = x^2 - y^2). since you have an i in there you'll wind up with -(i^2) which is (-(-1) or +1. So the complex number "disappears" and you're left with all real coefficients.
Just multiply it out, very carefully, be especially careful with + and - signs and order of operations.
Sometimes in a question like this they might just give you one of the complex roots and you'll have to know that they always come in conjugate pairs to come up with the other root. For example they might say find an equation wtih all real coefficients and with roots 3 and 2+i. You'll need an additional root, 2-i, to make it come out with all real coefficients.

Answer 2

The equation is the product of the factors, and the factors are the roots subtracted from the variable:

(x-3)*[x-(1+i)]*[x-(1-i)];

First multiply out the factors with imaginary terms (remember i*i=-1);
Their product is x^2 - (1 + i + 1 - i)*x + (1 + i)*(1 - i)=
x^2 - 2x + (1 + 1)= x^2 - 2x + 2;

Then multiply this times the remaining factor:

(x-3)*(x^2 - 2x + 2) = x^3 - 2x^2 + 2x - 3x^2 + 6x - 6 =

x^3 - 5x^2 + 8x - 6 = 0

Answer 3

Assuming the original expression is a polynomial and there are no repeating roots, then this should be equal to (x-3)(x - (1+i))(x - (1-i)) = 0. Multiply it out to find the original expression.

Start with (x - (1+i))(x - (1-i)). Using the FOIL method, this is

x^2 -(1+i)x - (1-i)x + (1+i)(1-i) =
x^2 - x - ix - x + ix + (1+i)(1-i) =
x^2 - 2x + (1+i)(1-i) =
x^2 - 2x + (1 - i^2) =
x^2 - 2x + (1 - (-1)) =
x^2 - 2x + 2

Now multiply this by the other factor, (x-3). When you mulitply and recombine the terms, you should get x^3 - 5x^2 + 8x - 6.

Answer 4

1st get the quadratic.
(x-1-i)(x-1+i)=0
x^2+x(-1-i) +x(-1+i)+(-1+i)(-1-i)=0
x^2+2x+((-1)^2-i^2))=0
x^2+2x=1-(-1)=0
x^2+2x+2=0
(x-3)(x^2+2x+2)=0
x^3+2x^2+2x
-3x^2-6x-6
x^3-x^2-4x-6=0