document.write( "Question 594467: 5x(x+1)(x-1)>0 \n" ); document.write( "
Algebra.Com's Answer #376742 by mamiya(56)\"\" \"About 
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5x(x+1)(x-1)>0\r
\n" ); document.write( "\n" ); document.write( " first, solve for 5x(x+1)(x-1)=0
\n" ); document.write( " 5x=0 x = 0
\n" ); document.write( "5x(x+1)(x-1) = 0 --> x+1 =0 --> x = -1
\n" ); document.write( " x-1 = 0 x = 1\r
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\n" ); document.write( "\n" ); document.write( "now, we have to test the function f(x)= 5x(x+1)(x-1) with its roots ( 0, -1, 1), by taking any numbers in all those intervals , plugging that in the function and putting the sign of the answer in the interval you took the number from. ( consider oo as infinity and -oo as minus infinity)\r
\n" ); document.write( "\n" ); document.write( " let's take, -2 from (-oo, -1) f(-2) = 5(-2)(-2+1)(-2-1) = -30
\n" ); document.write( " -1/2 from ( -1, 0) f(-1/2) = 5(-1/2)(-1/2-1)(-1/2+1)= + 15/8
\n" ); document.write( " 1/2 from (0, 1) f(1/2) = 5(1/2)(1/2-1)(1/2+1) = -15/8
\n" ); document.write( " 2 from ( 1, oo) f(2) = 5(2)(2-1)(2+1) = + 30
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\n" ); document.write( " we get this (-oo) ----- (-1) ++++ (0) ----(1) ++++ (oo) \r
\n" ); document.write( "\n" ); document.write( "looking at this , there are only two intervals satisfying that question. those intervals are (-1, 0) and (1, oo)\r
\n" ); document.write( "\n" ); document.write( "so the answer is (-1, 0)U(1, oo) \n" ); document.write( "
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