document.write( "Question 24416: The polynomial equation
\n" ); document.write( " x(x^2+4)(x^2-x-6)=0. Has how many real roots?\r
\n" ); document.write( "\n" ); document.write( "We are very rusty on our Algebra skills, and we are reviewing for a math placement exam, b/c we are returning back to college.\r
\n" ); document.write( "\n" ); document.write( "Thanks,
\n" ); document.write( "K \n" ); document.write( "

Algebra.Com's Answer #12993 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
You can apply the zero product principle to this problem: If $\"a%2Ab+=+0\"$ then either $\"a+=+0\"$ or $\"b+=+0\"$ or both.
\n" ); document.write( " $\"x%28x%5E2%2B4%29%28x%5E2-x-6%29+=+0\"$
\n" ); document.write( " $\"x+=+0\"$ and/or $\"x%5E2%2B4+=+0\"$ and/or $\"x%5E2-x-6+=+0\"$
\n" ); document.write( "One of the real roots is x = 0
\n" ); document.write( "If $\"x%5E2%2B4+=+0\"$ then $\"x%5E2+=+-4\"$ and x = +or- $\"sqrt%28-4%29\"$ so these two roots are:
\n" ); document.write( "x = 2i and x = -2i These are not real roots
\n" ); document.write( "If $\"x%5E2-x-6+=+0\"$ then $\"%28x%2B2%29%28x-3%29+=+0\"$ so $\"x+=+-2\"$ and $\"x+=+3\"$ These two roots are real.\r
\n" ); document.write( "\n" ); document.write( "The real roots are:
\n" ); document.write( "x = 0
\n" ); document.write( "x = -2
\n" ); document.write( "x = 3 \n" ); document.write( "

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