What is (3x-8)^4(x-1)^3(x+1)<= 0 ?

The solution to (3x-8)^4(x-1)^3(x+1)<= 0 is -1<= x<= 1\lor x= 8/3

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(3x-8)^4(x-1)^3(x+1)<= 0

(3 x - 8)^{4} (x - 1)^{3} (x + 1) \leq 0

- 1 \leq x \leq 1 or x = \frac{8}{3}

Interval Notation

[- 1, 1] \cup x = \frac{8}{3}

Decimal Notation

- 1 \leq x \leq 1 or x = 2.66666 \dots

Solution steps

(3 x - 8)^{4} (x - 1)^{3} (x + 1) \leq 0

Identify the intervals

- 1 \leq x \leq 1 or x = \frac{8}{3}

3 X - 6 + 4 X = 7 - 3 X + 5

(x + 4) (5 x + 9) - x^{2} + 16 = 0

- 52 a^{2} + 5119 a - 54421 = 40000

lo g_{81} (27)

z - 4.1 > - 8.6 + 2 z

What is (3x-8)^4(x-1)^3(x+1)<= 0 ?
The solution to (3x-8)^4(x-1)^3(x+1)<= 0 is -1<= x<= 1\lor x= 8/3