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

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

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

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

x = 0 or x > 1

Interval Notation

x = 0 \cup (1, \infty)

Solution steps

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

Rewrite in standard form

\frac{x ^{2} ( x ^{2} + 1 ) ( x - 2 ) ^{2}}{( 24 x ^{4} + 2304 x ^{2} + 4 ( 96 x ^{3} + 1536 x ) + 6144 ) ( x - 1 ) ^{3}} \geq 0

Factor

\frac{x ^{2} ( x ^{2} + 1 ) ( x - 2 ) ^{2}}{( 24 x ^{4} + 2304 x ^{2} + 4 ( 96 x ^{3} + 1536 x ) + 6144 ) ( x - 1 ) ^{3}} : \frac{x ^{2} ( x ^{2} + 1 ) ( x - 2 ) ^{2}}{24 ( x + 4 ) ^{4} ( x - 1 ) ^{3}}

\frac{x ^{2} ( x ^{2} + 1 ) ( x - 2 ) ^{2}}{24 ( x + 4 ) ^{4} ( x - 1 ) ^{3}} \geq 0

Multiply both sides by

24

\frac{24 x ^{2} ( x ^{2} + 1 ) ( x - 2 ) ^{2}}{24 ( x + 4 ) ^{4} ( x - 1 ) ^{3}} \geq 0 \cdot 24

Simplify

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

Identify the intervals

x = 0 or x > 1

12 - 3 x \geq 0

- 2 + x \leq 7

4 a^{2} - b^{2}

32 b^{2} = 2

4 \cdot 25

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