document.write( "Question 792243: P has degree 2 and zeros
\n" ); document.write( "1 + i\r
\n" ); document.write( "\n" ); document.write( "5
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\n" ); document.write( "1 − i\r
\n" ); document.write( "\n" ); document.write( "5
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Algebra.Com's Answer #479896 by josgarithmetic(39617) $\"\"$ $\"About$

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Recheck your problem description. If P is degree two, then you have TWO zeros; you cannot have zeros 1+i, 1-i, and 5 as the zeros. IF you have 1+i as a zero, then you ALSO have 1-i as a zero because complex zeros occur in P as \"Conjugate Pairs\", so you cannot also have 5 as a zero. If your zeros are supposed to be 1+i and 5, then you must also have 1-i; and this means P must be AT LEAST degree 3. \r
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\n" ); document.write( "\n" ); document.write( "For degree THREE, your question would work and have a solution, but not for degree TWO. \r
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\n" ); document.write( "\n" ); document.write( "You would, for degree THREE, have factors for P being $\"%28x-5%29\"$ , $\"%28x-%281%2Bi%29%29\"$ , and $\"%28x-%281-i%29%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Recheck your problem description from your book or source carefully, and then ask what you really want help with. \n" ); document.write( "

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