What is (x-4)^2(16-2x)>0 ?

The solution to (x-4)^2(16-2x)>0 is x<4\lor 4<x<8

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(x-4)^2(16-2x)>0

(x - 4)^{2} (16 - 2 x) > 0

x < 4 or 4 < x < 8

Interval Notation

(- \infty, 4) \cup (4, 8)

Solution steps

(x - 4)^{2} (16 - 2 x) > 0

Rewrite in standard form

- x^{3} + 16 x^{2} - 80 x + 128 > 0

Factor

- x^{3} + 16 x^{2} - 80 x + 128 : - (x - 4)^{2} (x - 8)

- (x - 4)^{2} (x - 8) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 4)^{2} (x - 8)) (- 1) < 0 \cdot (- 1)

Simplify

(x - 4)^{2} (x - 8) < 0

Identify the intervals

x < 4 or 4 < x < 8

a + b

- 6 x^{6} = 2 x - 9

\frac{7}{x + 3} > - 5

4 x - 3 (2 x + 1) \leq 4 (x - 3)

x^{2} - 6 x + 14 = 0