document.write( "Question 380012: Find the polynomial of lowest degree whose roots are 2i and 8i \n" ); document.write( "

Algebra.Com's Answer #269749 by richard1234(7193) $\"\"$ $\"About$

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Suppose our polynomial is P(x). If P(x) has roots 2i and 8i, then P(x) = (x-2i)(x-8i)Q(x), where Q(x) is an arbitrary polynomial. If Q(x) = 1 (to minimize the degree), then $\"P%28x%29+=+%28x-2i%29%28x-8i%29+=+x%5E2+-+10ix+-+16\"$ . The only issue is that we have a non-real coefficient, 10i.\r
\n" ); document.write( "\n" ); document.write( "Suppose we wanted to find a polynomial with only real coefficients.\r
\n" ); document.write( "\n" ); document.write( "I'll use a trial and error case: By Vieta's formulas, we want the sum and the product of the roots to be real numbers. Therefore I'll introduce the conjugates of the first two roots, -2i and -8i. Our polynomial becomes\r
\n" ); document.write( "\n" ); document.write( "(x-2i)(x+2i)(x-8i)(x+8i)\r
\n" ); document.write( "\n" ); document.write( "= $\"%28x%5E2+%2B+4%29%28x%5E2+%2B+64%29\"$
\n" ); document.write( "= $\"x%5E4+%2B+68x%5E2+%2B+256\"$ \r
\n" ); document.write( "\n" ); document.write( "Either answer could be correct, depending on whether the coefficients are assumed to be real. \n" ); document.write( "

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