zeros are i and (1+i)

degree of the polynomial 4
constant term 12

How do i do it?
(x+i)(x-i)(x-1-i)(x-1+i)=0 ?
Where does the 12 go?

Thanks

Where does the 12 go?

Substitute x = 0 and find out what the constant term is now. Then multiply the polynomial you have now by a factor that will make the constant term equal to 12.

To determine where the constant term of 12 goes in the polynomial, you can follow these steps:

1. Begin with the given zeros: i and (1+i). Since complex zeros occur in conjugate pairs, the conjugates of i and (1+i) are -i and (1-i), respectively. Therefore, your factors are: (x + i), (x - i), (x - 1 - i), and (x - 1 + i).

2. Expand the factors: Multiply all the factors together using the distributive property to obtain the polynomial in its factored form: (x + i)(x - i)(x - 1 - i)(x - 1 + i) = 0.

3. Simplify the factors: To simplify the factors, use the difference of squares formula to eliminate complex numbers. The difference of squares formula states that a^2 - b^2 = (a + b)(a - b). Applying this formula to the factors, we have: (x^2 - i^2)(x^2 - (1 + i)(1 - i)) = 0.

4. Simplify further: Simplify the factors by applying the difference of squares formula to eliminate complex numbers: (x^2 + 1)(x^2 - 1 - i^2) = 0.

5. Simplify the remaining imaginary numbers: Simplify the imaginary numbers within the factors: (x^2 + 1)(x^2 - 1 - (-1)) = 0.

6. Further simplify: Simplify further if possible: (x^2 + 1)(x^2 - 1 + 1) = 0.

7. Reduce the polynomial: Continue simplifying: (x^2 + 1)(x^2) = 0.

8. Multiply the constants: Distribute and multiply the constants: x^4 + x^2 = 0.

Now, observe that the constant term in this polynomial is 0, not 12 as mentioned in your question. If you indeed want the constant term to be 12, you would need to multiply the entire polynomial by a constant factor that will yield a constant term of 12 when x = 0.