Use the table of signs of linear factors to determine the solution set of the following: (x+4)(x-1)(x-3)<=0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the solution of: (x+4)(x-1)(x-3)<=0

(x+4)(x-1)(x-3) = 0 is true if either of the terms is equal to 0.

=> x = -4, x = 1 and x = 3

(x+4)(x-1)(x-3)<0 when either one of terms if negative or all three are negative.

(x+4) < 0, (x-1) > 0 and (x-3) > 0

=> x < -4, x > 1 and x > 3

This is not possible

x + 4 >0, x - 1 < 0 and x -3 >0

=> x > -4, x < 1 and x > 3

This is not possible

x + 4 >0, x - 1 >0 and x -3 <0

=> x > -4, x < 1 and x < 3

This gives the set of solutions (-4, 1)

x + 4 < 0, x - 1< 0 and x - 3 < 0

=> x > -4, x > 1 and x > 3

This gives the set of solutions (3, inf.)

The solution set of the inequality is [-4, 1] U [3, inf.)

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