Form a polynomial f(x) with real coeffiecients having the given degree and zeros.
degree 4, zeros 3, multiplicity 2; 6i
f(x)=a()
Can someone show how to do this. I have tried and do not have a clue. I am lost!!! So, please show work.
by the time you get to calculus, you should know how to use the roots to form a polynomial. If you tried, you should show what you have tried, and where you got stuck.
If a is a root of f(x)=0, (x-a) is a factor of f(x).
Since complex roots come in pairs, -6i is also a root, so
f(x) = (x-3)(x-3)(x-6i)(x+6i)
= (x-3)^2 (x^2+36)
saying you do not have a clue indicates either
a) you have woefully neglected your studies, or
b) you are just lazy and want someone else to do your work, or
(c) both of the above.
To form a polynomial with the given degree and zeros, we need to consider the fact that the zeros of a polynomial are the values of x for which the polynomial evaluates to zero.
Here's how you can proceed:
Step 1: Start by determining the factors of the polynomial based on the zeros given. In this case, we have two zeros: 3 with a multiplicity of 2, and 6i.
Step 2: For each zero, write a factor of the polynomial using the zero and its multiplicity.
For the zero 3 with multiplicity 2, the factor will be (x - 3)(x - 3) or (x - 3)^2.
For the zero 6i, the factor will be (x - 6i).
Step 3: Combine all the factors to form the polynomial. Since the coefficients of the polynomial are not specified, we'll use the coefficient 'a' to represent an arbitrary constant.
f(x) = a(x - 3)^2(x - 3)(x - 6i)
This is the polynomial with real coefficients and the given degree and zeros.
Note: Remember that imaginary zeros always come in conjugate pairs. Therefore, if 6i is a zero, then -6i must also be a zero. So, you could alternatively write the polynomial as:
f(x) = a(x - 3)^2(x - 3)(x - 6i)(x + 6i)
I hope this helps! Let me know if you have any further questions.