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

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

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

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

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

Interval Notation

[- 2, 1] \cup x = 2

Solution steps

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

Rewrite in standard form

x^{6} - 3 x^{5} - 2 x^{4} + 12 x^{3} - 8 x^{2} \leq 0

Factor

x^{6} - 3 x^{5} - 2 x^{4} + 12 x^{3} - 8 x^{2} : x^{2} (x - 1) (x - 2)^{2} (x + 2)

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

Identify the intervals

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

\frac{( 3 x + 4 )}{( 2 x - 1 )} < 2

4 x^{3} - 17 x^{2} + 21 x - 8 \geq 0

\frac{( - ( x - 1 ) - 2 )}{x} \leq 0

2 x \geq x^{2}

\frac{( x - 1 ) ( x - 2 ) ^{2}}{( x + 5 )} \geq 0

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