Question 634915: Find a polynomial with integer coefficients taht satisfies the given conditions. Q has degree 3 and zeros 0 and i. Found 2 solutions by Alan3354, jsmallt9:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find a polynomial with integer coefficients taht satisfies the given conditions. Q has degree 3 and zeros 0 and i.
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Complex solutions occur in conjugate pairs, so -i is also a solution
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--> (x-0)*(x-i)*(x+i) = 0
Not sure what the Q is about.
You can put this solution on YOUR website! The Fundamental Theorem of Algebra tells us that a polynomial with real coefficients and degree n, will have n zeros. Since integers are real numbers, our polynomial Q will have 3 zeros since its degree is 3.
But we were only given two zeros. We will need all three to get an answer. Another property of polynomials with real coefficients is that if a zero is complex, then that zero's complex conjugate will also be a zero. The standard form for complex numbers is:
a + bi
The complex conjugate of this would be
a - bi
or
a + (-bi)
In this problem you have been given a complex zero: i. In standard form this would be:
0 + i
So it complex conjugate:
0 - i (or just -i)
will also be a zero.
So now we have all three zeros: 0, i and -i.
To create our polynomial we will use this form:
Where "a" can be any non-zero real number we choose and the z's are our three zeros. Since we want Q to have integer coefficients then we should choose a non-zero integer for "a". (Since there are an infinite number of possible a's there are an infinite number of polynomials that will have our three zeros. This is why the problem says "Find a polynomial..." instead of "Find the polynomial...".
The simplest choice for "a" is 1. Using this for "a" and substituting our zeros in we get:
Now we simplify
Since what we have left is multiplication and since order doesn't matter when multiplying, I recommend that you start with multiplying the factors with the complex conjugate roots. There are two reasons for this:
The multiplication is easy because you can use the pattern to do it quickly. And...
The i's will disappear which will make the remaining multiplications easier.
So we will multiply the last two factors first, using the pattern:
Since this simplifies:
Multiplying by the x:
This is "a" polynomial with integer coefficients with the given zeros.
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