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

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

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

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

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

Interval Notation

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

Decimal Notation

x < - 1.41421 \dots or - 1 \leq x \leq 1 or x > 1.41421 \dots

Solution steps

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

Factor

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

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

Identify the intervals

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

\frac{1}{x} < 3

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

(x - 5) (x + 5) < 0

- 2 x^{2} + x - 1 > 0

x^{2} + 18 < 5 x

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