Find a polynomial function of lowest degree with rational coefficients that has (1+i) and 2 as some of it's zeros.
Hi bailey526,
Remember:
If 2 is a zero, then (x - 2) is a factor.
Imaginary zeros come in conjugate pairs, so 1 - i is a zero and 1 + i is a zero.
Therefore, (x - (1 - i)) is a factor and (x - (1 + i)) is a factor.
Set these factors to zero and expand to find your equation.
(x - 2)(x - (1 - i))(x - (1 + i)) = 0
It looks like you might not be familiar with the concepts involved here...?
Think about when you solved quadratic equations by factoring. You'd get a factored equation that looked something like (x - a)(x - b) = 0, and you'd solve by setting each factor equal to zero: x - a = 0 or x - b = 0, so x = a or x = b.
Now, you need to work backwards. Given that x = a and x = b are zeroes, they must have solved x - a = 0 and x - b = 0, so x - a and x - b must have been factors.
By multiplying these factors, you can find the quadratic from its zeroes.
Note, however, that 3(x - a)(x - b) has the same zeroes (since the "3" will be divided off). Note also that (x - a)^3 (x - b)^4 has the same zeroes, but is of higher degree.
To find "the polynomial of least degree" with a given set of zeroes, you don't over-duplicate like I just did above.
For worked examples of how this process works with polynomials (not just quadratics) and complex roots, try here.