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-1<x<4

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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 - 1 < x < 4

Interval Notation

(- \infty, - 2] \cup (- 1, 4)

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 )}{- ( x + 1 ) ( x - 4 )}

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x \leq - 2 or - 1 < x < 4

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

\frac{( 5 - 2 x )}{( 3 x ^{2} - 2 x - 16 )} < 1

(x^{2} - 4 x + 4) (1 - 2 x) (x^{2} + 3 x - 1) x > 0

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

\frac{( 3 x - 2 )}{( 2 x - 3 )} < 3

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-1<x<4