Could you help me with the following problem, I don't understand how to do it. Am I suppose to use the linear factorization theorem?
Find an nth degree polynomial function with real coefficients satisfying the given conditions.
1. n=3; 3 and i are zeros; f(2)=20
my answer choices are:
a. f(x)=-4^3+12x^2-4x+12
b.f(x)=-4^3+12x^2+4x-12
c.f(x)=4^3-12x^2-4x+12
d.f(x)=-4^3-12x^2+4x-12
I already gave you a function that worked. There are others that also work. Since this is multiple choice, just plug in the numbers 1, 2 and 3 into each equation and see which one satisfies the three conditions. (b) does not satisfy the f(1) = 0 condition, for example.
Yes, you are correct that the linear factorization theorem can be used to solve this problem. The theorem states that if a polynomial function has a zero x = a, then it can be factored into the form (x - a) times another polynomial.
To find the nth degree polynomial function that satisfies the given conditions, we need to find the linear factors given the zeros and then multiply those factors together.
In this case, we are given that the zeros are 3 and i. Since i is a complex number, it comes with its conjugate, -i. Therefore, the linear factors are (x - 3), (x - i), and (x + i).
To find the final polynomial function, we multiply these linear factors together:
(x - 3)(x - i)(x + i)
Expanding this expression, we get:
(x - 3)(x^2 - i^2)
Since i^2 is equal to -1, we can simplify further:
(x - 3)(x^2 + 1)
Now, we have a polynomial of degree 3. To find the specific coefficients, we can use the given condition, f(2) = 20.
Substituting x = 2 into our polynomial, we get:
(2 - 3)(2^2 + 1) = -1(4 + 1) = -1(5) = -5
We want the polynomial to evaluate to 20, so we need to multiply it by -4:
-4(-5) = 20
Therefore, the final polynomial function is:
f(x) = -4(x - 3)(x^2 + 1)
By comparing this expression with the answer choices you provided, we can see that the correct answer is:
b. f(x) = -4^3 + 12x^2 + 4x - 12