form a polynomial f(x) with real coefficients having the given degree and zeros.
degree: 4; zeros: -1, 2, and 1-2i.
I got an exam tomorrow, i would appreciate any kind of help, thank you.
You are given 2 real roots and one complex.
As I noted before, complex numbers come in pairs,
so there has to be another one, 1+2i
so we have
f(x) = (x+1)(x-2)(x-1-2i)(x-1+2i)
multiply the last two brackets, the i's will drop out
thank you again, life saver
To form a polynomial with the given degree and zeros, you can use the fact that if a is a root of a polynomial, then (x - a) is a factor of the polynomial.
Given that the zeros are -1, 2, and 1-2i, we know that the factors of the polynomial will be (x + 1), (x - 2), and (x - (1-2i)).
First, let's find the factor for the complex zero 1-2i.
Since it is a complex zero, its conjugate will also be a zero. Therefore, the conjugate of 1-2i is 1+2i.
So, the factors for the complex zero 1-2i are (x - (1-2i)) and (x - (1+2i)).
Now, we can multiply these factors together to get the polynomial:
(x + 1)(x - 2)(x - (1-2i))(x - (1+2i)) = 0
To simplify this, we can use the fact that for complex numbers a+bi, a-bi is also a root. So instead of multiplying out the complex factors separately, we can combine them:
[(x - (1-2i))(x - (1+2i))] = (x - (1-2i))(x - (1+2i)) = (x - 1 + 2i)(x - 1 - 2i) = (x - 1)^2 - (2i)^2
Simplifying further:
(x - 1)^2 - (2i)^2 = x^2 - 2x + 1 - 4i^2 = x^2 - 2x + 1 - 4(-1) = x^2 - 2x + 1 + 4 = x^2 - 2x + 5
So, the polynomial with degree 4 and zeros -1, 2, and 1-2i is:
f(x) = (x + 1)(x - 2)(x^2 - 2x + 5)
Therefore, f(x) = (x + 1)(x - 2)(x^2 - 2x + 5).