What is (2x+1)(x-2)(3x-4)<= 0 ?

The solution to (2x+1)(x-2)(3x-4)<= 0 is x<=-1/2 \lor 4/3 <= x<= 2

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(2x+1)(x-2)(3x-4)<= 0

(2 x + 1) (x - 2) (3 x - 4) \leq 0

x \leq - \frac{1}{2} or \frac{4}{3} \leq x \leq 2

Interval Notation

(- \infty, - \frac{1}{2}] \cup [\frac{4}{3}, 2]

Decimal Notation

x \leq - 0.5 or 1.33333 \dots \leq x \leq 2

Solution steps

(2 x + 1) (x - 2) (3 x - 4) \leq 0

Identify the intervals

x \leq - \frac{1}{2} or \frac{4}{3} \leq x \leq 2

16 x^{4} - (y - 2 z)^{2}

2 x^{2} - 4 x y + 2 y^{2}

\frac{i - 2}{i - 1}

x y + 7 y - 3 x - 21

- 1 (1 + 7 x) - 6 (- 7 - x) = 36

What is (2x+1)(x-2)(3x-4)<= 0 ?
The solution to (2x+1)(x-2)(3x-4)<= 0 is x<=-1/2 \lor 4/3 <= x<= 2