document.write( "Question 370375: Factor the following polynomials completely making use of the given zero
\n" ); document.write( "G(x)=x^3-(1-i)X^2-(8-i)x+(12-6i);2-i is a zero \n" ); document.write( "

Algebra.Com's Answer #264206 by jsmallt9(3759) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"G%28x%29=x%5E3-%281-i%29x%5E2-%288-i%29x%2B%2812-6i%29\"$
\n" ); document.write( "If 2-i is a zero, then (x-(2-i)) or (x - 2 + i) is a factor of G(x). We need to find the other factor and for this we can use Synthetic Division:
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document.write( "2-i |   1  -1+i  -8+i  12-6i\r\n" );
document.write( "-----       2-i   2-i -12+6i\r\n" );
document.write( "       ----------------------\r\n" );
document.write( "        1   1    -6     0\r\n" );
document.write( "

\n" ); document.write( "Te remainder is zero so 2-i is indeed a factor. And the rest of that row of numbers, 1 1 -6, tells us the other factor: $\"1x%5E2+%2B+1x+-+6\"$ . So now
\n" ); document.write( " $\"G%28x%29+=+%28x-2%2Bi%29%28x%5E2%2Bx-6%29\"$
\n" ); document.write( "The second factor is a trinomial that is easily factored giving:
\n" ); document.write( " $\"G%28x%29+=+%28x-2%2Bi%29%28x%2B3%29%28x-2%29%29\"$
\n" ); document.write( "G(x) is now fully factored. \n" ); document.write( "

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