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

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

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

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

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x \leq - 2 or x = 1

x^{2} > - 4

x^{2} > - 2

48 + 2 x - x^{2} \geq 0

\frac{5}{6 ( x + 2 ) - 4} < \frac{x}{3 + 12}

3 \cdot (x - 1)^{2} - 12 \leq 6 x^{2} - (3 + x)^{2}

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