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

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

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

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

x \leq - 1 or x = 1 or x \geq 4

Interval Notation

(- \infty, - 1] \cup x = 1 \cup [4, \infty)

Solution steps

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

Identify the intervals

x \leq - 1 or x = 1 or x \geq 4

4 x^{3} - 3 x^{2} - 16 x + 12 < 0

\frac{( m + 1 )}{m} \leq 0

\frac{( x ^{2} + 6 x - 7 )}{( x ^{2} + 1 )} \leq 2

2 x^{2} + 2 x + 1 < \frac{15}{( x ^{2} + x + 1 )}

\frac{2 + 3}{( x + 1 )} > \frac{2}{x}

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