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

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

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

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

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

x = - 1, x = \frac{1}{2}, x = - 3, x = 3

Solution steps

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

Factor

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

(x + 1) (2 x - 1) (x + 3) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or 2 x - 1 = 0 or x + 3 = 0 or x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

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

Solve

x + 3 = 0 : x = - 3

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 1, x = \frac{1}{2}, x = - 3, x = 3

7 x^{9} = (7 x^{10}) - (7 x)

x, 27 x^{4} + 60 x^{2} = 32

8 x^{3} = 0

0 = x^{4} - 18 x^{2} + 32

x (x - 5)^{3} - 14 (x - 5)^{2} = 0

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