What is -5(x-5)(x+4)^2(x-7)>= 0 ?

The solution to -5(x-5)(x+4)^2(x-7)>= 0 is x=-4\lor 5<= x<= 7

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-5(x-5)(x+4)^2(x-7)>= 0

- 5 (x - 5) (x + 4)^{2} (x - 7) \geq 0

x = - 4 or 5 \leq x \leq 7

Interval Notation

x = - 4 \cup [5, 7]

Solution steps

- 5 (x - 5) (x + 4)^{2} (x - 7) \geq 0

Rewrite in standard form

- x^{4} + 4 x^{3} + 45 x^{2} - 88 x - 560 \geq 0

Factor

- x^{4} + 4 x^{3} + 45 x^{2} - 88 x - 560 : - (x + 4)^{2} (x - 5) (x - 7)

- (x + 4)^{2} (x - 5) (x - 7) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x + 4)^{2} (x - 5) (x - 7)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x + 4)^{2} (x - 5) (x - 7) \leq 0

Identify the intervals

x = - 4 or 5 \leq x \leq 7

- 5 ∣ 1 - 2 x ∣ + 6 = - 54

4 (- 8 x + 12) = - 26 - 32 x

5 x^{2} = 125

y - 9 < 11

5 + 8 ∣ - 10 n - 2 ∣ = 101

What is -5(x-5)(x+4)^2(x-7)>= 0 ?
The solution to -5(x-5)(x+4)^2(x-7)>= 0 is x=-4\lor 5<= x<= 7