Find a polynomial function f(x), with real coefficients, that has 1 and 3+2i as zeros, and such that f(-1)=2 (Multiply out and simplify your answer)
f(x) = a(x-1)(x-(3+2i))(x-(3-2i))
= a(x-1)(x^2-6x+13)
= a(x^3-7x^2+19x-13
f(-1) = a(-1-7-19-13) = 40a = 2
so, a = 1/20
To find the polynomial function with the given zeros, we can use the fact that if a polynomial has a root at a particular value, then it will have a factor of (x - root).
So, the polynomial function can be written as:
f(x) = (x - 1)(x - (3 + 2i))(x - (3 - 2i))
Multiplying this out and simplifying, we have:
f(x) = (x - 1)(x^2 - 6x + 13)
Expanding the first term, we get:
f(x) = (x^3 - 6x^2 + 13x) - (x^2 - 6x + 13)
Combining like terms, we have:
f(x) = x^3 - 7x^2 + 19x - 13
Therefore, the polynomial function f(x) with real coefficients that has 1 and 3+2i as zeros and such that f(-1) = 2 is:
f(x) = x^3 - 7x^2 + 19x - 13.
To find a polynomial function f(x) with the given zeros and f(-1) = 2, we can use the fact that if a complex number is a zero, then its conjugate is also a zero.
Given that the zeros are 1 and 3+2i, we know that the conjugate of 3+2i is 3-2i.
To find the polynomial function, we can use the fact that if a number is a zero, then x - zero = 0.
So, our polynomial function will be (x - 1)(x - (3+2i))(x - (3-2i)).
Multiplying this out, we get:
= (x - 1)(x - 3 - 2i)(x - 3 + 2i)
Now, let's simplify this expression:
= (x - 1)((x - 3)^2 - (2i)^2)
= (x - 1)(x^2 - 6x + 9 - 4i^2)
= (x - 1)(x^2 - 6x + 9 + 4)
= (x - 1)(x^2 - 6x + 13)
= x^3 - 6x^2 + 13x - x^2 + 6x - 13
= x^3 - 7x^2 + 19x - 13
Therefore, the polynomial function f(x) with the given zeros and f(-1) = 2 is:
f(x) = x^3 - 7x^2 + 19x - 13.