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

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

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

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

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

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

Decimal Notation

x = - 1, x = - 0.5, x = 3

Solution steps

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

Move

3

to the left side

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

Move

x

to the left side

2 x^{3} - 4 x^{2} - 8 x - 3 = - x^{2}

Move

x^{2}

to the left side

2 x^{3} - 3 x^{2} - 8 x - 3 = 0

Factor

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

(x + 1) (2 x + 1) (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

Solve

x + 1 = 0 : x = - 1

Solve

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

Solve

x - 3 = 0 : x = 3

The solutions are

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

x^{4} - 14 x^{2} + 49 = 0

x^{6} - 23 x^{4} + 4 x^{2} = 0

y^{3} + 3 y^{2} = 4 y + 12

x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6 = 0

8 x^{3} - 1 = 0

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