information is given about the polynomial f(x) whose coefficients are real numbers. find the real zeros of f:
degree 4; zeros: i, 3+i
complex roots always come in conjugate pairs
so if i is a root so is -i, and if 3+i is a root, so is 3-i
so f(x) = (x^2+1)(x - (3+i))(x - (3-i)
= (x^2 + 1)((x^2 + 6x + 10)
There are no real zeros.
am sorry i wrote in incoreectly forgive me, it was suppose to say: information is given about the polynomial f(x) whose coefficients are real numbers. find the remaining zeros of f: degree 4; zeros: i, 3+i, sorry again and thank you for taking the time to explain to me and solve the problem, god bless you always:)
To find the real zeros of the polynomial f(x), we need to consider that complex zeros always appear in conjugate pairs. Since the given zeros are i (which is complex) and 3+i (also complex), we know that the conjugate of i is -i and the conjugate of 3+i is 3-i.
To find the real zeros, we equate the polynomial to zero by using the complex conjugates of the given zeros:
(x - i)(x + i)(x - (3+i))(x - (3-i)) = 0
Expanding the equation, we get:
(x^2 - i^2)(x^2 - (3+i)(3-i)) = 0
Simplifying further, we have:
(x^2 + 1)(x^2 - (9 - i^2)) = 0
Since i^2 is equal to -1, the equation becomes:
(x^2 + 1)(x^2 - 8) = 0
Now we can solve for x:
1) x^2 + 1 = 0
To solve this equation, we subtract 1 from both sides:
x^2 = -1
Taking the square root of both sides:
x = ±i
2) x^2 - 8 = 0
To solve this equation, we add 8 to both sides:
x^2 = 8
Taking the square root of both sides:
x = ±√8
Simplifying the second equation further:
x = ±√(4 * 2)
x = ±2√2
In summary, the real zeros of the polynomial f(x) are: ±2√2.