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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(((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{0.5}{x - x ^{2} - 1} < 0

(x - 1)^{3} < 4 (x - 1)

(x + 2)^{2} > 9

\frac{1}{( x - 2 )} + \frac{1}{( x - 1 )} > \frac{1}{x}

h, h_{2} \cdot s + s \cdot o_{2} - > h_{2} \cdot o + s

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)