Question 819796: Write a third degree polynomial function y=P(x) with rational coefficients so that P(x)=0 has the given roots.
-3, i
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! A third degree polynomial has three roots. With only two roots given it would seem that something is missing. The "trick" is to know that if a polynomial with real (rational is a subset of real) coefficients has a complex root, a + bi, then its complex conjugate, a - bi, will also be a root.
So if i is a root, so will its conjugate. "i", in a + bi form, is 0 + 1i. So its conjugate would be 0 - 1i, or just -i.
Now we have all three roots: -3, i and -i. And when a number is a root of a polynomial then (x - that number) is a factor of the polynomial. So we can write P(x) in factored form:
Simplifying the factors we get:
All that is left is to multiply this out. Tip: If there are complex roots, then the multiplying is easier if you multiply the factors of theses roots first. So we will start by multiplying the last two factors. We can use the  pattern to multiply them quickly:
which simplifies as follows:
Since  by definition:
Now we multiply the remaining factors:
Simplifying:
Reordering the terms to put the polynomial in standard, highest exponent to lowest exponent order:
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