What is (x-7)(x+4)^2(x-1)>= 0 ?

The solution to (x-7)(x+4)^2(x-1)>= 0 is x = 7

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(x-7)(x+4)^2(x-1)>= 0

(x - 7) (x + 4)^{2} (x - 1) \geq 0

x \leq 1 or x \geq 7

Interval Notation

(- \infty, 1] \cup [7, \infty)

Solution steps

(x - 7) (x + 4)^{2} (x - 1) \geq 0

Identify the intervals

x \leq 1 or x \geq 7

- \frac{1}{- 3}

\frac{13 !}{( 13 - 4 ) ! \cdot 4 !}

\frac{x}{3} - x = \frac{- x + 19}{5}

v^{2} = \frac{3.9388}{0.0452}

4 - \frac{3}{2} x = 16

What is (x-7)(x+4)^2(x-1)>= 0 ?
The solution to (x-7)(x+4)^2(x-1)>= 0 is x<= 1\lor x>= 7