What are the solutions to the equation 2x^3-7x^2+4x+1=0 ?

The solutions to the equation 2x^3-7x^2+4x+1=0 are x=1,x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

Find the zeros of 2x^3-7x^2+4x+1=0

The zeros of 2x^3-7x^2+4x+1=0 are x=1,x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4

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2 x^{3} - 7 x^{2} + 4 x + 1 = 0

x = 1, x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

Decimal Notation

x = 1, x = 2.68614 \dots, x = - 0.18614 \dots

Solution steps

2 x^{3} - 7 x^{2} + 4 x + 1 = 0

Factor

2 x^{3} - 7 x^{2} + 4 x + 1 : (x - 1) (2 x^{2} - 5 x - 1)

(x - 1) (2 x^{2} - 5 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 2 x^{2} - 5 x - 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x^{2} - 5 x - 1 = 0 : x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

The solutions are

x = 1, x = \frac{5 + 33}{4}, x = \frac{5 - 33}{4}

2 t^{3} - 3 t^{2} - 2 t - 42 = 0

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

- x^{3} + 10 x^{2} - 33 x + 36 = 0

3 x^{3} - 24 x^{2} + 21 x = 0

x^{3} - 6 x = 3 x^{2} - 8

What are the solutions to the equation 2x^3-7x^2+4x+1=0 ?
The solutions to the equation 2x^3-7x^2+4x+1=0 are x=1,x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4
Find the zeros of 2x^3-7x^2+4x+1=0
The zeros of 2x^3-7x^2+4x+1=0 are x=1,x=(5+sqrt(33))/4 ,x=(5-sqrt(33))/4