What are the solutions to the equation 4x^3-27x^2-11x-1=0 ?

The solutions to the equation 4x^3-27x^2-11x-1=0 are x=-1/4 ,x=(7+sqrt(53))/2 ,x=(7-sqrt(53))/2

Find the zeros of 4x^3-27x^2-11x-1=0

The zeros of 4x^3-27x^2-11x-1=0 are x=-1/4 ,x=(7+sqrt(53))/2 ,x=(7-sqrt(53))/2

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4 x^{3} - 27 x^{2} - 11 x - 1 = 0

x = - \frac{1}{4}, x = \frac{7 + 53}{2}, x = \frac{7 - 53}{2}

Decimal Notation

x = - 0.25, x = 7.14005 \dots, x = - 0.14005 \dots

Solution steps

4 x^{3} - 27 x^{2} - 11 x - 1 = 0

Factor

4 x^{3} - 27 x^{2} - 11 x - 1 : (4 x + 1) (x^{2} - 7 x - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

x^{2} - 7 x - 1 = 0 : x = \frac{7 + 53}{2}, x = \frac{7 - 53}{2}

The solutions are

x = - \frac{1}{4}, x = \frac{7 + 53}{2}, x = \frac{7 - 53}{2}

(1 - m)^{3} = \frac{1}{3}

x^{1} + 2 x^{2} + x^{3} = - 1

4 x^{3} - 54 x^{2} + 180 x - 125 = 0

2 k^{3} - 8 k^{2} + 2 k + 1 = 0

b^{4} - 6 b^{3} - 5 b^{2} + 42 b + 40 = 0

What are the solutions to the equation 4x^3-27x^2-11x-1=0 ?
The solutions to the equation 4x^3-27x^2-11x-1=0 are x=-1/4 ,x=(7+sqrt(53))/2 ,x=(7-sqrt(53))/2
Find the zeros of 4x^3-27x^2-11x-1=0
The zeros of 4x^3-27x^2-11x-1=0 are x=-1/4 ,x=(7+sqrt(53))/2 ,x=(7-sqrt(53))/2