document.write( "Question 723008: (y^3-y^2-3y-9)/(y+1)
\n" ); document.write( "the ^ is suppose to mean its a exponent so y exponent 3 and y exponent 2
\n" ); document.write( "**Sorry about that \n" ); document.write( "

Algebra.Com's Answer #443027 by KMST(5328) $\"\"$ $\"About$

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The exponents are understood.
\n" ); document.write( "y^3 is the usual way to type $\"y%5E3\"$ and is y^2 the usual way to type $\"y%5E2\"$
\n" ); document.write( "What is not clear is what to do with that rational expression.
\n" ); document.write( "We can factor or divide. Maybe one of those two choices is what you wanted.
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\n" ); document.write( "FACTORING:
\n" ); document.write( " $\"%28y%5E3-y%5E2-3y-9%29%2F%28y%2B1%29=%28y-3%29%28y%5E2%2B2y%2B3%29%2F%28y%2B1%29\"$
\n" ); document.write( "and we see that it cannot be simplified.
\n" ); document.write( "The possible rational zeros of $\"y%5E3-y%5E2-3y-9\"$ are -9, -3, -1, 1, 3, and 9.
\n" ); document.write( "It was easy to try -3, -1, 1, and 3.
\n" ); document.write( "Since 3 was a zero $\"y-3\"$ had to be a factor.
\n" ); document.write( "Dividing by $\"%28y-3%29\"$ we get $\"y%5E2%2B2y%2B3\"$ and
\n" ); document.write( " $\"y%5E2%2B2y%2B3\"$ cannot be factored further because it has no zeros.
\n" ); document.write( "(There are no real solutions to $\"y%5E2%2B2y%2B3=0\"$ ).
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\n" ); document.write( "DIVIDING:
\n" ); document.write( " $\"%28y%5E3-y%5E2-3y-9%29%2F%28y%2B1%29=y%5E2-2y-1-8%2F%28y%2B1%29\"$
\n" ); document.write( "We get $\"y%5E2-2y-1\"$ for a quotient and a remainder of $\"-8\"$ . \n" ); document.write( "

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