Question 677975: Find a polynomial f(x) of degree 3 with real coefficients and the following zeros.
-4, -1, i
Thanks in advance for your help
Found 2 solutions by swincher4391, stanbon:
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! I think you may have meant one of two things. Either the degree is supposed to be 4, or -1 was supposed to be -i. As you may know, imaginary roots come in conjugate pairs, so with i, we are guaranteed -i to be a root as well. So if -4,-1,i are roots then -i is a root as well, which gives us degree 4.
So I'll solve it both ways.
If you meant -4,-1,i with degree 4, that's
(x+4)(x+1)(x-i)(x+i)
[x^2+5x+4][x^2+1] = x^4 + 5x^3 + 5x^2 + 5x + 4
If you meant -4,-i,i then that's
(x+4)(x-i)(x+i)
(x+4)(x^2+1) = x^3 + 4x^2+ x + 4
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Find a polynomial f(x) of degree 3 with real coefficients and the following zeros.
-4, -1, i
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Real Number coefficients implies that -i must also be a zero.
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Ans: f(x) = (x+4)(x+1)(x-i)(x+i)
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f(x) = (x^2+5x+4)(x^2+1) = x^4 + 5x^3 + 4x^2 + x^2 + 5x + 4
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= x^4 + 5x^3 + 5x^2 + 5x + 4
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Cheers,
Stan H.
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