What is (2x-7)^4(x-1)^3(x+1)<= 0 ?

The solution to (2x-7)^4(x-1)^3(x+1)<= 0 is -1<= x<= 1\lor x= 7/2

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(2x-7)^4(x-1)^3(x+1)<= 0

(2 x - 7)^{4} (x - 1)^{3} (x + 1) \leq 0

- 1 \leq x \leq 1 or x = \frac{7}{2}

Interval Notation

[- 1, 1] \cup x = \frac{7}{2}

Decimal Notation

- 1 \leq x \leq 1 or x = 3.5

Solution steps

(2 x - 7)^{4} (x - 1)^{3} (x + 1) \leq 0

Identify the intervals

- 1 \leq x \leq 1 or x = \frac{7}{2}

3 = \frac{n}{6}

x^{- \frac{7}{2}} + 2 x^{- \frac{5}{2}} + x^{- \frac{3}{2}}

15 a x + 3 a y

\frac{- 3 + 11 i}{4 + 7 i}

∣ 3 x - 5 ∣ \geq 1

What is (2x-7)^4(x-1)^3(x+1)<= 0 ?
The solution to (2x-7)^4(x-1)^3(x+1)<= 0 is -1<= x<= 1\lor x= 7/2