What is (x-5)^2(32-4x)>0 ?

The solution to (x-5)^2(32-4x)>0 is x<5\lor 5<x<8

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(x-5)^2(32-4x)>0

(x - 5)^{2} (32 - 4 x) > 0

x < 5 or 5 < x < 8

Interval Notation

(- \infty, 5) \cup (5, 8)

Solution steps

(x - 5)^{2} (32 - 4 x) > 0

Rewrite in standard form

- x^{3} + 18 x^{2} - 105 x + 200 > 0

Factor

- x^{3} + 18 x^{2} - 105 x + 200 : - (x - 5)^{2} (x - 8)

- (x - 5)^{2} (x - 8) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 5)^{2} (x - 8)) (- 1) < 0 \cdot (- 1)

Simplify

(x - 5)^{2} (x - 8) < 0

Identify the intervals

x < 5 or 5 < x < 8

\leq 0 = 2 x + 11

8^{- 4}

8 i^{2} (3 - 5 i)

- 4 x^{2} - 20 x + 10 = 0

r^{2} - 4 r + 2 = 0