What is x^4*(x-1)^3*(x+2)^2*(x-5)*(x+6)^2<= 0 ?

The solution to x^4*(x-1)^3*(x+2)^2*(x-5)*(x+6)^2<= 0 is x=-6\lor x=-2\lor x=0\lor 1<= x<= 5

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x^4(x-1)^3(x+2)^2(x-5)(x+6)^2<= 0

x^{4} \cdot (x - 1)^{3} \cdot (x + 2)^{2} \cdot (x - 5) \cdot (x + 6)^{2} \leq 0

x = - 6 or x = - 2 or x = 0 or 1 \leq x \leq 5

Interval Notation

x = - 6 \cup x = - 2 \cup x = 0 \cup [1, 5]

Solution steps

x^{4} (x - 1)^{3} (x + 2)^{2} (x - 5) (x + 6)^{2} \leq 0

Identify the intervals

x = - 6 or x = - 2 or x = 0 or 1 \leq x \leq 5

x (x + 3) < 5 - x

[x]^{2} + 5 [x] - 6 < 0

x^{2} + 2 x - 8 =< 6 x - x^{2} + 1

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

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

What is x^4*(x-1)^3*(x+2)^2*(x-5)*(x+6)^2<= 0 ?
The solution to x^4*(x-1)^3*(x+2)^2*(x-5)*(x+6)^2<= 0 is x=-6\lor x=-2\lor x=0\lor 1<= x<= 5