Find a fourth-degree polynomial with integer coefficients that has zeros 4i and −1, with −1 a zero of multiplicity 2. (Use x for the variable.)
can somebody help me do this?
If 4i is a zero, then -4i must also be a zero, since complex zeros come in conjugate pairs.
a possible polynomial is
f(x) = (x-4i)(x+4i)(x+1)^2
a general solution would be
f(x) = a(x-4i)(x+4i)(x+1)^2 , where a is a non-zero integer.
If necessary, your can expand this.
To find a fourth-degree polynomial with integer coefficients that has zeros 4i and -1 (with -1 a zero of multiplicity 2), we can start by writing out the factors of the polynomial.
Since the zeros are 4i and -1 (with a multiplicity of 2), the factors of the polynomial are as follows:
(x - 4i)(x + 4i)(x + 1)(x + 1)
Now let's simplify these factors.
The factor (x - 4i)(x + 4i) simplifies to:
(x - 4i)(x + 4i) = x^2 + 4ix - 4ix - (4i)^2
= x^2 + 16i^2
= x^2 + 16(-1)
= x^2 - 16
The factor (x + 1)(x + 1) simplifies to:
(x + 1)(x + 1) = (x + 1)^2
= x^2 + 2x + 1
Now let's multiply these factors together:
(x^2 - 16)(x^2 + 2x + 1)
Expanding the multiplication:
(x^2 - 16)(x^2 + 2x + 1) = x^4 + 2x^3 + x^2 - 16x^2 - 32x - 16 + x^2 + 2x + 1
Combining like terms:
x^4 + 2x^3 - 16x^2 - 32x + x^2 + 2x - 16 + 1
Simplifying further:
x^4 + 2x^3 - 15x^2 - 28x - 15
Therefore, the fourth-degree polynomial with integer coefficients that has zeros 4i and -1 (with -1 a zero of multiplicity 2) is:
x^4 + 2x^3 - 15x^2 - 28x - 15
Yes, I can help you with that! To find the polynomial with the given zeros, we need to consider the complex zeros and the multiplicity of the real zero.
First, let's start with the complex zeros. You are given two complex zeros: 4i and -4i. Complex zeros always come in conjugate pairs, so if 4i is a zero, then -4i must also be a zero.
Since complex zeros occur in pairs, we can write two factors for the polynomial:
(x - 4i) and (x + 4i)
Next, let's move on to the real zero with multiplicity. In this case, -1 has a multiplicity of 2. This means that it is a zero twice.
Therefore, we can write another factor for the polynomial:
(x + 1)^2
Now, we just need to combine these factors together to form the polynomial. To do that, we multiply all the factors together:
P(x) = (x - 4i)(x + 4i)(x + 1)^2
But there is a way to simplify this expression. Since the complex zeros are conjugates, their product will always be a difference of squares. In other words, (x - 4i)(x + 4i) is equal to (x^2 + 16).
Therefore, the final polynomial with integer coefficients is:
P(x) = (x^2 + 16)(x + 1)^2
And that's it! You have found the fourth-degree polynomial with integer coefficients that has zeros 4i, -4i, and -1 with a multiplicity of 2.