find a 4th degree polynomial with integer coefficients that has zeros 3i and -1 with -1 a zero of multiplicity 2
y = (x-3i)(x+3i)(x+1)^2 = (x^2+9)(x+1)^2
To find a 4th degree polynomial with integer coefficients that has zeros 3i, -1 (with a multiplicity of 2), and some unknown zero, we can use the fact that complex roots occur in conjugate pairs.
Given the zeros, we can write the polynomial as:
(x - 3i)(x + 3i)(x + 1)(x + 1)
By expanding this expression, we get:
[(x^2 - (3i)^2)(x + 1)](x + 1)
[(x^2 + 9)(x + 1)](x + 1)
Expanding further:
(x^2 + 9)(x + 1)(x + 1)
(x^2 + 9)(x^2 + 2x + 1)
x^4 + 2x^3 + x^2 + 9x^2 + 18x + 9
x^4 + 2x^3 + 10x^2 + 18x + 9
Therefore, the 4th degree polynomial with integer coefficients that has the given zeros is:
f(x) = x^4 + 2x^3 + 10x^2 + 18x + 9.
To find a 4th degree polynomial with the given zeros, we can use the fact that if a complex number is a zero of a polynomial with real coefficients, then its conjugate is also a zero.
Given that the zeros are 3i, -1 (with multiplicity 2), and their conjugates -3i and -1, respectively, we can write the polynomial as:
(x - 3i)(x + 3i)(x + 1)(x + 1)
To simplify the expression, we can use the fact that (a + b)(a - b) = a^2 - b^2:
[(x)^2 - (3i)^2][(x + 1)^2]
Simplifying further:
[x^2 - (9i^2)] * [(x + 1)^2]
Since i^2 = -1:
[x^2 + 9](x + 1)^2
Expanding further:
(x^2 + 9)(x^2 + 2x + 1)
Finally, expanding the binomial (x^2 + 2x + 1)(x^2 + 9):
x^4 + 2x^3 + x^2 + 9x^2 + 18x + 9
Combining like terms:
x^4 + 2x^3 + 10x^2 + 18x + 9
Therefore, the 4th degree polynomial with the given zeros is:
f(x) = x^4 + 2x^3 + 10x^2 + 18x + 9.