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

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

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

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

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

Interval Notation

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

Decimal Notation

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

Solution steps

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

Identify the intervals

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

\frac{21 a ^{2}}{2 b} \cdot \frac{4 b ^{2}}{3 a ^{2}}

lo g_{7} (7)

3 x^{2} + 17 x + 10

- x^{2} + 6 x - 5 = 0

5 x^{2} + 15 x + 1 = 0

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