What is (x+1)(x^2-5x+6)<= 0 ?

The solution to (x+1)(x^2-5x+6)<= 0 is x<=-1\lor 2<= x<= 3

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(x+1)(x^2-5x+6)<= 0

(x + 1) (x^{2} - 5 x + 6) \leq 0

x \leq - 1 or 2 \leq x \leq 3

Interval Notation

(- \infty, - 1] \cup [2, 3]

Solution steps

(x + 1) (x^{2} - 5 x + 6) \leq 0

Factor

(x + 1) (x^{2} - 5 x + 6) : (x + 1) (x - 2) (x - 3)

(x + 1) (x - 2) (x - 3) \leq 0

Identify the intervals

x \leq - 1 or 2 \leq x \leq 3

9 = x 2

128 c^{6}

x^{3} - 2 x^{2} - 14 x + 3

(- \frac{4}{3})^{3}

9 m^{2} + 30 m + 25

What is (x+1)(x^2-5x+6)<= 0 ?
The solution to (x+1)(x^2-5x+6)<= 0 is x<=-1\lor 2<= x<= 3