# SOLUTION: Solve (x+5)(x-9)(x+3)>0, the solution set is {x| }

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 Click here to see ALL problems on Polynomials-and-rational-expressions Question 629459: Solve (x+5)(x-9)(x+3)>0, the solution set is {x| } Answer by Edwin McCravy(8893)   (Show Source): You can put this solution on YOUR website!```(x+5)(x-9)(x+3) > 0 The critical numbers are numbers which when substituted for x causes the left side either equal to 0 or undefined. These critical numbers are -5, 9, and -3 Mark those with open circles on the number line: ----------o-----o-----------------------------------o------------ -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 We choose a number less than -5, say -6, and substitute it in the inequality: (x+5)(x-9)(x+3) > 0 (-6+5)(-6-9)(-6+3) > 0 (-1)(-15)(-3) > 0 -45 > 0 That's false, so we do not shade the number line left of -5 We choose a number between -5 and -3, say -4, and substitute it in the inequality: (x+5)(x-9)(x+3) > 0 (-4+5)(-4-9)(-4+3) > 0 (1)(-13)(-1) > 0 13 > 0 That's true, so we do shade the number line between -5 and -3: ----------o=====o-----------------------------------o------------ -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 We choose a number between -3 and 9, say 0, and substitute it in the inequality: (x+5)(x-9)(x+3) > 0 (0+5)(0-9)(0+3) > 0 (5)(-9)(3) > 0 -135 > 0 That's false, so we do not shade the number line between -3 and 9: We choose a number greater than 9, say 10, and substitute it in the inequality: (x+5)(x-9)(x+3) > 0 (10+5)(10-9)(10+3) > 0 (15)(1)(13) > 0 195 > 0 That's true, so we do shade the number line right of 9 ----------o=====o-----------------------------------o============> -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Answer: {x| -5 < x < -3 OR x > 9 } Edwin```