document.write( "Question 31602: simplify/Reduce
\n" ); document.write( "16 4
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\n" ); document.write( "x^2+2x-3 x+3 \n" ); document.write( "
Algebra.Com's Answer #18261 by mbarugel(146)\"\" \"About 
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Hello!
\n" ); document.write( "We can start by finding the roots of the quadratic equation in the denominator of the 1st term:\r
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"1x%5E2%2B2x%2B-3+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%282%29%5E2-4%2A1%2A-3=16\".
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\n" ); document.write( " Discriminant d=16 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-2%2B-sqrt%28+16+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%282%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+1\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%282%29-sqrt%28+16+%29%29%2F2%5C1+=+-3\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B2x%2B-3\" can be factored:
\n" ); document.write( " \"1x%5E2%2B2x%2B-3+=+%28x-1%29%2A%28x--3%29\"
\n" ); document.write( " Again, the answer is: 1, -3.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-3+%29\"

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\n" ); document.write( "\n" ); document.write( "The roots are 1 and -3. This implies that that quadratic equation can be written as:\r
\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B2x-3+=+%28x-1%29%28x+-%28-3%29%29+=+%28x-1%29%28x%2B3%29\"\r
\n" ); document.write( "\n" ); document.write( "So your expression becomes:\r
\n" ); document.write( "\n" ); document.write( "\"16%2F%28%28x-1%29%28x%2B3%29%29+-+4%2F%28x%2B3%29\"\r
\n" ); document.write( "\n" ); document.write( "But since 4 is a factor of 16 and (x+3) is a factor of (x-1)(x+3), then we can take \"4%2F%28x%2B3%29\" as a common factor, getting:\r
\n" ); document.write( "\n" ); document.write( "\"%284%2F%28x%2B3%29%29%28+4%2F%28x-1%29+-+1%29\"\r
\n" ); document.write( "\n" ); document.write( "Furthermore, we can add the fractions inside the parenthesis, getting:\r
\n" ); document.write( "\n" ); document.write( "\"%284%2F%28x%2B3%29%29%28%28+4+-+%28x-1%29%29%2F%28x-1%29%29\"
\n" ); document.write( "\"%284%2F%28x%2B3%29%29%28%28+5-x%29%2F%28x-1%29%29\"\r
\n" ); document.write( "\n" ); document.write( "And that's as far as we can go.\r
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\n" ); document.write( "\n" ); document.write( "I hope this helps!
\n" ); document.write( "Get more answers at Online Math Answers.com!
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