write a polynomial function of least degree that has real coefficeints the given zeros and a leading coefficient of 1. the problem is 5,2i,-2i
*i think you mean the ZEROS / ROOTS are 5, 2i and -2i.
if so, you can get them by multiplying and expanding,
(x-5)(x-2i)(x+2i)
note that 1 = sqrt(-1)
but an easier method would be, to start with the roots 2i and -2i. recall that q quadratic equation follows the formula,
x^2 - (sum of roots)*x + (product of roots)
thus, substituting 2i and -2i:
x^2 - (2i - 2i)*x + (2i)*(-2i)
x^2 - (0)*x - 4*(i^2)
x^2 - 4*(-1)
x^2 + 4
now, we multiply this by (x-5):
(x-5)(x^2 + 4)
x^3 - 5x^2 + 4x - 20
hope this helps~ :)
To create a polynomial function with the given zeros, we can use the fact that the complex zeros occur in conjugate pairs.
The zeros are 5, 2i, and -2i.
First, let's consider the real zero, which is 5. To express this as a factor of the polynomial, we subtract it from x: (x - 5).
Next, let's consider the complex zeros. The complex zeros occur in conjugate pairs, which means if one zero is 2i, the other is -2i. To express these as factors of the polynomial, we subtract them from x: (x - 2i) and (x + 2i).
To get the polynomial function, we multiply these factors together:
(x - 5)(x - 2i)(x + 2i)
Now, let's simplify this expression.
When multiplying the complex conjugates (x - 2i) and (x + 2i), we use the difference of squares formula: (a + b)(a - b) = a^2 - b^2. In this case, a = x and b = 2i.
(x - 5)(x^2 - (2i)^2)
(x - 5)(x^2 - 4i^2)
(x - 5)(x^2 - 4(-1))
(x - 5)(x^2 + 4)
Therefore, the polynomial function of least degree with the given zeros and a leading coefficient of 1 is: f(x) = (x - 5)(x^2 + 4).