document.write( "Question 1178287: Use the given zero to find all other zeros.
\n" ); document.write( "Zero:2i
\n" ); document.write( "Equation: F(x)=2x^3+3x^2+8x+12
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Algebra.Com's Answer #807520 by MathTherapy(10555)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Use the given zero to find all other zeros.
\n" ); document.write( "Zero:2i
\n" ); document.write( "Equation: F(x)=2x^3+3x^2+8x+12
\n" ); document.write( "Thanks in advance!
\n" ); document.write( "
\"matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+2x%5E3+%2B+3x%5E2+%2B+8x+%2B+12%29\".
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document.write( "It's NOT as complex as the other 2 people state. 
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document.write( "You don't even need to use the 2 roots that're given (2i and its conjugate, - 2i) because when EACH binomial, from left to right, is factored, we get: 
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document.write( "You MIGHT be able to recognize \"x%5E2+%2B+4\" as the EXPANDED form of the given zero, 2i and its conjugate.
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document.write( "With that being said, the other factor, as seen above, is 2x + 3, which gives us: 2x + 3 = 0 ====> 2x = - 3, and ultimately, the other zero, or: \"highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+highlight%28-+3%2F2%29%29%29\" \n" );
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