What are the solutions to the equation 2x^4-3x^3-7x^2+12x-4=0 ?

The solutions to the equation 2x^4-3x^3-7x^2+12x-4=0 are x=1,x= 1/2 ,x=-2,x=2

Find the zeros of 2x^4-3x^3-7x^2+12x-4=0

The zeros of 2x^4-3x^3-7x^2+12x-4=0 are x=1,x= 1/2 ,x=-2,x=2

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2 x^{4} - 3 x^{3} - 7 x^{2} + 12 x - 4 = 0

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

Decimal Notation

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

Solution steps

2 x^{4} - 3 x^{3} - 7 x^{2} + 12 x - 4 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

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

\frac{1}{4} x^{4} = 4

x^{3} - 3 x^{2} - 7 x + 21 = 0

(9 x - 3)^{4} - 2 = 0

x = c x (x - 1)^{2}

0 = 2 x^{3} + 3 x^{2} - 18 x - 27

What are the solutions to the equation 2x^4-3x^3-7x^2+12x-4=0 ?
The solutions to the equation 2x^4-3x^3-7x^2+12x-4=0 are x=1,x= 1/2 ,x=-2,x=2
Find the zeros of 2x^4-3x^3-7x^2+12x-4=0
The zeros of 2x^4-3x^3-7x^2+12x-4=0 are x=1,x= 1/2 ,x=-2,x=2