SOLUTION: Explain whether or not there exists a third-degree polynomial with integer coefficients that has no real zeroes.

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Question 1026249: Explain whether or not there exists a third-degree polynomial with integer coefficients that has no real zeroes.
Found 2 solutions by robertb, josgarithmetic:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
There does NOT exist a 3rd degree polynomial with integer coefficients that has no real zeroes. The fact that if a pure complex number (one that contains "i") is a zero then guarantees its conjugate is also a zero implies that the third zero has to be without the imaginary unit i. (This is necessary, because the presence of the integer coefficients would force the absence of i for the 3rd zero.)

Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
If you found this example, y=%28x%2B6%29%28x%5E2%2B5%29, you would understand that this has only a single real zero, while the other TWO zeros are not real. The degree of the equation, if as a function, is odd, specifically here, being 3. NOT even, but odd. Could you imagine any such y=%28x-a%29%28x%5E2%2Bb%29 for which the root x-a would be not Real? Assuming a and b are Real...