A polynomial function f(x) has degree 6 and has real coefficients. It is given that 3, 2, 11−3i, and 11+28i are roots of f(x). What is the sum of all the roots of f(x)?
the other two roots are 11+3i and 11-28i
So, just add them all up. 3+2+22+22 = 49
To find the sum of all the roots of the polynomial function, we first need to determine the remaining two roots since it is given that there are six roots in total.
Since the given polynomial has real coefficients and 11−3i is a root, we know that its conjugate, 11+3i, must also be a root. Similarly, the conjugate of 11+28i would be 11-28i.
Therefore, the six roots of the polynomial function f(x) are: 3, 2, 11−3i, 11+3i, 11−28i, and 11+28i.
To find the sum of the roots, we can simply add up all the values:
3 + 2 + 11−3i + 11+3i + 11−28i + 11+28i
Now, notice that the imaginary parts (the i) cancel out when adding the complex conjugate pairs. This means that:
(11−3i + 11+3i) and (11−28i + 11+28i) both simplify to 22.
Therefore, the sum of all the roots is:
3 + 2 + 22 + 22 = 49.
Hence, the sum of all the roots of f(x) is 49.