What is x^4+2x^3-7x^2-8x+12<= 0 ?

The solution to x^4+2x^3-7x^2-8x+12<= 0 is -3<= x<=-2\lor 1<= x<= 2

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x^4+2x^3-7x^2-8x+12<= 0

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 \leq 0

- 3 \leq x \leq - 2 or 1 \leq x \leq 2

Interval Notation

[- 3, - 2] \cup [1, 2]

Solution steps

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 \leq 0

Factor

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 : (x - 1) (x + 3) (x + 2) (x - 2)

(x - 1) (x + 3) (x + 2) (x - 2) \leq 0

Identify the intervals

- 3 \leq x \leq - 2 or 1 \leq x \leq 2

12 x^{2} + 7 x - 12 > 0

x^{2} + 4 x + 2 x + 8

2 x + 9 + 3 x + x = 5 x + 9

lo g_{27} (\frac{1}{9})

x^{3} + 3 x^{2} - 3 x - 9

What is x^4+2x^3-7x^2-8x+12<= 0 ?
The solution to x^4+2x^3-7x^2-8x+12<= 0 is -3<= x<=-2\lor 1<= x<= 2