Find a polynomial function having the following zeros:
– 5, 3 (multiplicity 2), 2 + i, 2 – i; degree 5.
let's first do the quadratic which would give us roots of 2 ± i
(x - (2+i))(x - (2-i)) = .....
= x^2 - 4x + 5
so we have
f(x) = (x+5)(x-3)(x-3)(x^2 - 4x + 5)
which, if expanded will give us a 5th degree polynomial.
To find a polynomial function with the given zeros, we can use the factored form of a polynomial function. The factored form of a polynomial is written as the product of its linear factors.
Given that the zeros are -5, 3 (with multiplicity 2), 2 + i, and 2 - i, we can write the factored form of the polynomial as follows:
(x + 5)(x - 3)(x - 3)(x - (2 + i))(x - (2 - i))
Now let's simplify this expression by using conjugate pairs. The term (x - (2 + i))(x - (2 - i)) represents a pair of complex conjugate zeros.
(x + 5)(x - 3)(x - 3)[(x - 2) - i][(x - 2) + i]
The conjugate pair can be written as (a - bi) and (a + bi), where a = 2 and b = 1. Expanding this expression, we get:
(x + 5)(x - 3)(x - 3)(x - 2 - i)(x - 2 + i)
Now let's simplify it further:
(x + 5)(x - 3)^2(x - 2 - i)(x - 2 + i)
Expanding the squared term, we have:
(x + 5)(x^2 - 6x + 9)(x - 2 - i)(x - 2 + i)
Multiplying the binomials, we get:
(x + 5)(x^2 - 6x + 9)(x^2 - 4x + 4 + i^2)
Since i^2 = -1, the expression becomes:
(x + 5)(x^2 - 6x + 9)(x^2 - 4x + 4 - 1)
Now let's simplify it further:
(x + 5)(x^2 - 6x + 9)(x^2 - 4x + 3)
Finally, let's multiply all the terms together to get the polynomial function:
f(x) = (x + 5)(x^2 - 6x + 9)(x^2 - 4x + 3)
Therefore, the polynomial function with the given zeros is f(x) = (x + 5)(x^2 - 6x + 9)(x^2 - 4x + 3) with a degree of 5.