What are the solutions to the equation 2x^3-7x^2-6x+8=0 ?

The solutions to the equation 2x^3-7x^2-6x+8=0 are x=4,x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4

Find the zeros of 2x^3-7x^2-6x+8=0

The zeros of 2x^3-7x^2-6x+8=0 are x=4,x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4

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2 x^{3} - 7 x^{2} - 6 x + 8 = 0

x = 4, x = \frac{- 1 + 17}{4}, x = \frac{- 1 - 17}{4}

Decimal

x = 4, x = 0.78077 \dots, x = - 1.28077 \dots

Solution steps

2 x^{3} - 7 x^{2} - 6 x + 8 = 0

Factor

2 x^{3} - 7 x^{2} - 6 x + 8 : (x - 4) (2 x^{2} + x - 2)

(x - 4) (2 x^{2} + x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 4 = 0 : x = 4

Solve

2 x^{2} + x - 2 = 0 : x = \frac{- 1 + 17}{4}, x = \frac{- 1 - 17}{4}

The solutions are

x = 4, x = \frac{- 1 + 17}{4}, x = \frac{- 1 - 17}{4}

x, x^{4} - 6 x^{2} + 12 = 7

25 x^{4} - 61 x^{2} + 36 = 0

y = x^{3}, x

x^{3} = 49

x^{3} = 48

What are the solutions to the equation 2x^3-7x^2-6x+8=0 ?
The solutions to the equation 2x^3-7x^2-6x+8=0 are x=4,x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4
Find the zeros of 2x^3-7x^2-6x+8=0
The zeros of 2x^3-7x^2-6x+8=0 are x=4,x=(-1+sqrt(17))/4 ,x=(-1-sqrt(17))/4