document.write( "Question 202835: The equation of a polynomial with real coefficients that has the roots 2 and 3(imaginary i) is?\r
\n" ); document.write( "\n" ); document.write( "a. x^3-2x^2+9x-18\r
\n" ); document.write( "\n" ); document.write( "b. x^3-2x^2-12x-18\r
\n" ); document.write( "\n" ); document.write( "c. x^3+2x^2-9x-18\r
\n" ); document.write( "\n" ); document.write( "d. x^3-2x^2+9x-20 \n" ); document.write( "

Algebra.Com's Answer #153093 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let's call the polynomial P(x). If 2 and 3i are roots of P(x) then P(x) must be:
\n" ); document.write( " $\"P%28x%29+=+%28x+-2%29%28x%5E2+%2B+9%29\"$
\n" ); document.write( "If this isn't immediately clear, think about it. If 2 is a root of P(x) then, by the very definition of a root, when x = 2, P(x) = 0. Can you see that if x = 2 in the equation above, that P(x) = 0? And similarly, can you see that if x = 3i (or -3i) that P(x) is zero (because $\"x%5E2+%2B+9+=+0\"$ if x = 3i (or -3i)?

\n" ); document.write( "Now all we need to do is multiply out P(x) to see which answer is correct. we should get:
\n" ); document.write( " $\"x%5E3+-+2x%5E2+%2B+9x+-18\"$ so (a) is correct. \n" ); document.write( "

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