find a third degree polynomial function with real coefficients -2+i and -4 zeros
To find a third degree polynomial function with real coefficients and the given complex zeros, we know that complex zeros always come in conjugate pairs. Since we have -2 + i as a zero, its conjugate -2 - i is also a zero.
Therefore, the three zeros of the polynomial function are: -2 + i, -2 - i, and -4.
To find the polynomial function, we can use the zero product property. The equation for the polynomial function can be written as follows:
(x - (-2 + i)) * (x - (-2 - i)) * (x - (-4)) = 0
Simplifying this equation, we have:
(x + 2 - i) * (x + 2 + i) * (x + 4) = 0
Expanding this equation further:
(x + 2 - i) * (x + 2 + i) * (x + 4) = (x^2 + 2x + ix + 2x + 4 + 2i - ix - 2i - i^2) * (x + 4)
(x + 2 - i) * (x + 2 + i) * (x + 4) = (x^2 + 4x + 4 + ix + 2i - ix - 2i + 1) * (x + 4)
(x + 2 - i) * (x + 2 + i) * (x + 4) = (x^2 + 4x + 5) * (x + 4)
Multiplying these factors together, we get:
(x + 2 - i) * (x + 2 + i) * (x + 4) = x^3 + 6x^2 + 13x + 20
Therefore, the third degree polynomial function with real coefficients and having -2 + i, -2 - i, and -4 as zeros is:
f(x) = x^3 + 6x^2 + 13x + 20
To find a third degree polynomial function with real coefficients and given zeros, consider the complex conjugate property of roots.
Since the coefficients are real, the complex conjugate of -2+i will also be a root of the polynomial. Hence, the roots are -2+i, its conjugate -2-i, and -4.
To find the polynomial, we can use the fact that if r is a root, then (x - r) is a factor of the polynomial. Therefore, the factors are (x - (-2+i)), (x - (-2-i)), and (x - (-4)).
Simplifying these factors, we have:
(x + 2 - i), (x + 2 + i), and (x + 4).
To obtain the polynomial, we multiply these factors together:
(x + 2 - i)(x + 2 + i)(x + 4).
Expanding this expression, we get:
(x^2 + 4x + 2ix + 2x + 4 + 2i - ix - 2i - i^2)(x + 4).
Simplifying, we have:
(x^2 + 6x + 4 + 2ix - ix - i^2)(x + 4).
Since i^2 = -1, the expression becomes:
(x^2 + 6x + 4 + x + 2i)(x + 4).
Further simplifying, we get:
(x^2 + 7x + 4 + 2i)(x + 4).
Expanding again, we obtain:
x^3 + 7x^2 + 4x + 2ix + 4x + 28 + 16i.
Combining like terms, the polynomial is:
x^3 + 7x^2 + 8x + 28 + (2i + 16i)x.
Therefore, a third-degree polynomial function with real coefficients and zeros -2+i, -2-i, and -4 is:
f(x) = x^3 + 7x^2 + 8x + 28 + 18ix.
check your typing
Your conditions make no sense.
Did you mean -4 is a zero and -2+i is a zero ?
if so, then we can do this:
complex roots always come in conjugate pairs
so if -2+i is a zero, then -2-i is also a zero
using the sum and product properties
sum of those two roots = -2+i + -2-i = -4
product of those two roots = (-2+i)(-2-i)
= 4 - i^2 = 5
so the quadratic for the two complex roots is
x^2 +4x + 5
then your cubic would be
y = (x+4)(x^2 + 4x + 5)
expand if needed