What are the solutions to the equation 2x^4-3x^3-4x^2+3x+2=0 ?

The solutions to the equation 2x^4-3x^3-4x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=2

Find the zeros of 2x^4-3x^3-4x^2+3x+2=0

The zeros of 2x^4-3x^3-4x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=2

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2 x^{4} - 3 x^{3} - 4 x^{2} + 3 x + 2 = 0

x = - 1, x = 1, x = - \frac{1}{2}, x = 2

Decimal Notation

x = - 1, x = 1, x = - 0.5, x = 2

Solution steps

2 x^{4} - 3 x^{3} - 4 x^{2} + 3 x + 2 = 0

Factor

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

(x + 1) (x - 1) (2 x + 1) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 1 = 0 or 2 x + 1 = 0 or x - 2 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

x - 2 = 0 : x = 2

The solutions are

x = - 1, x = 1, x = - \frac{1}{2}, x = 2

2 x^{3} + (- 7) x^{2} + 6 x + 5 = 0

b^{3} = 1331

x^{3} - 3.0909 x + 2.0909

- 4 a^{3} + 12 a^{2} - 12 a + \frac{7}{2}

12 x^{3} + 24 x^{2} - 36 x = 0

What are the solutions to the equation 2x^4-3x^3-4x^2+3x+2=0 ?
The solutions to the equation 2x^4-3x^3-4x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=2
Find the zeros of 2x^4-3x^3-4x^2+3x+2=0
The zeros of 2x^4-3x^3-4x^2+3x+2=0 are x=-1,x=1,x=-1/2 ,x=2