What is (((x+2)(x^2-2x+1)))/((-4+3x-2x^2))>= 0 ?

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

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(((x+2)(x^2-2x+1)))/((-4+3x-2x^2))>= 0

\frac{( ( x + 2 ) ( x ^{2} - 2 x + 1 ) )}{( - 4 + 3 x - 2 x ^{2} )} \geq 0

x \leq - 2 or x = 1

Interval Notation

(- \infty, - 2] \cup x = 1

Solution steps

\frac{( x + 2 ) ( x ^{2} - 2 x + 1 )}{- 4 + 3 x - 2 x ^{2}} \geq 0

Factor

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

\frac{( x - 1 ) ^{2} ( x + 2 )}{- ( 4 - 3 x + 2 x ^{2} )} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

\frac{( x - 1 ) ^{2} ( x + 2 )}{4 - 3 x + 2 x ^{2}} \leq 0

Identify the intervals

x \leq - 2 or x = 1

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

x < \frac{1}{x}

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

\frac{x}{x - 1} > \frac{1}{x}

\frac{( x + 5 )}{( x + 1 )} > 0

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