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

The solution to (x^2-3x+2)(x^3-3x^2)(4-x^2) = 3

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

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) \leq 0

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

Interval Notation

[- 2, 1] \cup x = 2 \cup [3, \infty)

Solution steps

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) \leq 0

Factor

(x^{2} - 3 x + 2) (x^{3} - 3 x^{2}) (4 - x^{2}) : - x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2)

- x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2)) (- 1) \geq 0 \cdot (- 1)

Simplify

x^{2} (x - 2)^{2} (x - 1) (x - 3) (x + 2) \geq 0

Identify the intervals

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

9 x^{2} - 6 < 19

\frac{( 3 x )}{( x - 8 )} < 0

x^{4} \cdot (x - 1)^{3} \cdot (x + 2)^{2} \cdot (x - 5) \cdot (x + 6)^{2} \leq 0

x (x + 3) < 5 - x

[x]^{2} + 5 [x] - 6 < 0

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