document.write( "Question 1027930: find nth degree given n=3;3&4i are zeros; f(1)=68
\n" ); document.write( "68=nth degree of((1)^3-(1)^2+16(1)-16) \n" ); document.write( "

Algebra.Com's Answer #643146 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
The assumption is that the coefficients of the cubic polynomial are real so the complex conjugate is also a root.
\n" ); document.write( " $\"f%28x%29=a%28x-3%29%28x-4i%29%28x%2B4i%29\"$
\n" ); document.write( " $\"f%28x%29=a%28x-3%29%28x%5E2%2B16%29\"$
\n" ); document.write( "Now use the point to solve for $\"a\"$ .
\n" ); document.write( " $\"f%281%29=a%281-3%29%281%5E2-16%29\"$
\n" ); document.write( " $\"68=a%28-2%29%28-15%29\"$
\n" ); document.write( " $\"a=68%2F30\"$
\n" ); document.write( " $\"a=34%2F15\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( " $\"f%28x%29=%2834%2F15%29%28x-3%29%28x%5E2%2B16%29\"$
\n" ); document.write( " \n" ); document.write( "

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