document.write( "Question 133139: Find a third degree polynomial f(x), which has roots 2, 1-2i and which has y-intercept 2 (ie, f(0) =2). \n" ); document.write( "

Algebra.Com's Answer #97354 by vleith(2983) $\"\"$ $\"About$

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You are given two root. One of those roots is imaginary. So there is another imaginary root.\r
\n" ); document.write( "\n" ); document.write( "roots are 2, 1-2i, 1+2i\r
\n" ); document.write( "\n" ); document.write( " $\"f%28x%29+=+%28x-2%29+%28x+-+%281-2i%29%29+%28x+-+%281%2B2i%29+%29\"$
\n" ); document.write( "multiply to get
\n" ); document.write( " $\"+f%28x%29+=+%28x-2%29%28x%5E2+-2x+%2B3%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "What is f(0)?
\n" ); document.write( " $\"f%280%29+=+%28-2%29%283%29\"$ = -6\r
\n" ); document.write( "\n" ); document.write( "We want f(0) to be 2. So we need to divide our function by -3.\r
\n" ); document.write( "\n" ); document.write( " $\"f%28x%29+=+-%28%28x-2%29%28x%5E2+-2x+%2B3%29%29%2F3+\"$ \n" ); document.write( "

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