Find Polynomial of lowest degree w/ zeros

A polynomial of lowest degree with real coefficients which has 2, 31/2 , 2i, and 3i as zeroes would be of degree

For this one I had no clue, I thought it was 4, but it was 6. What's the rule here?

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Originally Posted by nandokommando

A polynomial of lowest degree with real coefficients which has 2, 31/2 , 2i, and 3i as zeroes would be of degree

For this one I had no clue, I thought it was 4, but it was 6. What's the rule here?

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Hi nandokommando,

For any polynomial function, if an imaginary number is a zero of that function, its conjugate is also a zero.

Zeros = {2, 31/2, +2i, -2i, +3i, -3i}

Quote:

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Originally Posted by nandokommando

A polynomial of lowest degree with real coefficients which has 2, 31/2 , 2i, and 3i as zeroes would be of degree

For this one I had no clue, I thought it was 4, but it was 6. What's the rule here?

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As masters said, if a polynomial with real coefficients has a+ bi as zero it must also have it's complement, a- bi, as zero. A polynomial with real coefficients have 2, 31/2, 2i and 3i as zeros must also have -2i and -3i as zeros and so is of the form a(x- 2)(x- 31/2)(x- 2i)(x- 3i)(x+ 2i)(x+ 3i) and is degree 6.