document.write( "Question 6627: What is the possible number of imaginary zeros of $\"+f%28x%29+=+2x%5E3+-+4x%5E2+%2B+8x+-+3+\"$ ? \n" ); document.write( "

Algebra.Com's Answer #3613 by khwang(438) $\"\"$ $\"About$

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$\"+f%28x%29+=+2x%5E3+-+4x%5E2+%2B+8x+-+3+\"$ is a polynomial of third degree (odd)
\n" ); document.write( " in real coefficients.
\n" ); document.write( " Hence $\"+f%28x%29+=+2x%5E3+-+4x%5E2+%2B+8x+-+3+\"$ = 0 must have one real
\n" ); document.write( " or three real roots.\r
\n" ); document.write( "\n" ); document.write( " When it has only one real root, the other two are complex conjugate
\n" ); document.write( " number.
\n" ); document.write( " Hence, the possible number of imaginary zeros is 2 or 0.
\n" ); document.write( " [In fact, the complex roots appear always in pairs]\r
\n" ); document.write( "\n" ); document.write( " In fact, f(0) = -3 < 0 and f(1) = 3 > 0,so there is a zero of f
\n" ); document.write( " between 0 and 1. Also, since f'(x) = 6x^2 -8x + 8 = 2(3x^2 -4x + 4) =0
\n" ); document.write( " has no real roots, so f has only one real zero and two complex zeros.\r
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\n" ); document.write( "\n" ); document.write( " Kenny
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