Determine a polynomial function of degree 3 with real coefficients whose zeros are
ƒ{2, 1+i .
don't grok the ƒ{ part, but as for the rest, since complex roots occur in conjugate pairs,
p(x) = (x-2)(x-(1+i))(x-(1-i))
= (x-2)((x-1)-i)((x-1)+i)
= (x-2)(x^2-2x+1 + 1)
...
To determine a polynomial function with these zeros, we can use the concept of conjugate pairs.
Since the zeros are given as 2 and 1+i, we know that the conjugate of 1+i is 1-i. Therefore, the zeros are 2, 1+i, and 1-i.
To find the polynomial function, we can start by writing out the factors based on the zeros:
(x - 2)(x - (1+i))(x - (1-i))
Next, we can simplify the factors using the conjugate pairs:
(x - 2)(x - 1-i)(x - 1+i)
Expanding this expression, we get:
(x - 2)((x - 1) - i)((x - 1) + i)
Multiplying the factors together, we have:
(x - 2)((x - 1)^2 - i^2)
Simplifying further, we have:
(x - 2)(x^2 - 2x + 1 + 1)
Expanding again:
(x - 2)(x^2 - 2x + 2)
Finally, we distribute to get the polynomial function of degree 3:
f(x) = (x^3 - 2x^2 + 2x) + (-2x^2 + 4x - 4)
Simplifying:
f(x) = x^3 - 4x^2 + 6x - 4
Therefore, the polynomial function of degree 3 with real coefficients whose zeros are 2, 1+i, and 1-i is f(x) = x^3 - 4x^2 + 6x - 4.