document.write( "Question 470988: The solve and graph the solution set (x +2)(x-20)(x+1)>0 {x|_} \n" ); document.write( "

Algebra.Com's Answer #323074 by lwsshak3(11628) $\"\"$ $\"About$

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The solve and graph the solution set (x +2)(x-20)(x+1)>0
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\n" ); document.write( "f(x)= (x +2)(x-20)(x+1)>0
\n" ); document.write( "Draw a number line with zeros on it:\r
\n" ); document.write( "\n" ); document.write( "-∞..........-2.......... -1...........20..........∞ (zeros on number line)
\n" ); document.write( "-∞.....-.....0......+.....0.....-......0....+.....∞ (sign of (f(x))
\n" ); document.write( "Solution is to find the sign of f(x) within intervals between zeros on the number line.
\n" ); document.write( "Here is the procedure:
\n" ); document.write( "Starting from the right side,
\n" ); document.write( "For x>20, f(x)>0
\n" ); document.write( "For -1 < x< 20, f(x)<0
\n" ); document.write( "For -2 < x< -1, f(x)>0
\n" ); document.write( "For -2 < x< -1, f(x)<0
\n" ); document.write( "For -∞ < x< -2, f(x)>0
\n" ); document.write( "You can notice that the sign switches every time we go thru a zero. This happens when the zero is an odd multiplicity (1, 3. 5, etc) Since zeros are of multiplicity 1, sign of f(x) switches every time we go thru a zero. When zeros are of even mulltiplicity (2, 4, 6, etc),The sign of f(x) does not change as we go thru the zero on the number line. Once you have the correct number line showing signs of f(x), you have the solution.
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\n" ); document.write( "For given problem:
\n" ); document.write( "solution:(-2,-1) U (20,∞) \n" ); document.write( "

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