document.write( "Question 827428: find all the zeros of the equation x^4-3x^3+6x^2+2x-60=0 given that 1+3i is a zero \n" ); document.write( "
Algebra.Com's Answer #498719 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "\r\n" );
document.write( "x4-3x³+6x²+2x-60 = 0\r\n" );
document.write( "\r\n" );
document.write( "Use synthetic division:\r\n" );
document.write( "\r\n" );
document.write( "1+3i | 1  -3      6     2      -60\r\n" );
document.write( "     |     1+3i -11-3i  4-18i   60\r\n" );
document.write( "       1  -2+3i  -5-3i  6-18i    0\r\n" );
document.write( "\r\n" );
document.write( "So the first factorization is:\r\n" );
document.write( "\r\n" );
document.write( "[x-(1+3i)][x³+(-2+3i)x²+(-5-3i)x+(6-18i)] = 0\r\n" );
document.write( "\r\n" );
document.write( "Since 1+3i is a solution, so is its conjugate 1-3i\r\n" );
document.write( "\r\n" );
document.write( "1-3i | 1  -2+3i  -5-3i  6-18i\r\n" );
document.write( "     |     1-3i  -1+3i -6+18i\r\n" );
document.write( "       1  -1     -6\r\n" );
document.write( "\r\n" );
document.write( "The second factorization is:\r\n" );
document.write( "\r\n" );
document.write( "[x-(1+3i)][x-(1-3i)](x²-x-6) = 0\r\n" );
document.write( "\r\n" );
document.write( "The third and final factorization is:\r\n" );
document.write( "\r\n" );
document.write( "[x-(1+3i)][x-(1-3i)](x-3)(x+2) = 0\r\n" );
document.write( "\r\n" );
document.write( "x-(1+3i)=0\r\n" );
document.write( "       x=1+3i\r\n" );
document.write( "\r\n" );
document.write( "x-(1-3i)=0\r\n" );
document.write( "       x=1-3i\r\n" );
document.write( "\r\n" );
document.write( "     x-3=0\r\n" );
document.write( "       x=3\r\n" );
document.write( "\r\n" );
document.write( "     x+2=0\r\n" );
document.write( "       x=-2\r\n" );
document.write( "\r\n" );
document.write( "The four solutions are 1+3i, 1-3i, 3, -2\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );