What are the solutions to the equation 4x^3-5x+2=24 ?

The solutions to the equation 4x^3-5x+2=24 are x=2,x=-1+i(sqrt(7))/2 ,x=-1-i(sqrt(7))/2

Find the zeros of 4x^3-5x+2=24

The zeros of 4x^3-5x+2=24 are x=2,x=-1+i(sqrt(7))/2 ,x=-1-i(sqrt(7))/2

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4 x^{3} - 5 x + 2 = 24

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

Solution steps

4 x^{3} - 5 x + 2 = 24

Move

24

to the left side

4 x^{3} - 5 x - 22 = 0

Factor

4 x^{3} - 5 x - 22 : (x - 2) (4 x^{2} + 8 x + 11)

(x - 2) (4 x^{2} + 8 x + 11) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or 4 x^{2} + 8 x + 11 = 0

Solve

x - 2 = 0 : x = 2

Solve

4 x^{2} + 8 x + 11 = 0 : x = - 1 + i \frac{7}{2}, x = - 1 - i \frac{7}{2}

The solutions are

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

2 x^{3} - 6 x^{2} + 13 x - 14 = 0

x^{3} = - 1 - i

7 x^{6} + 35 x^{4} = 42 x^{5}

- 2 x (- x^{2} + 12) = 0

5 x + 21 = 10 x^{19}

What are the solutions to the equation 4x^3-5x+2=24 ?
The solutions to the equation 4x^3-5x+2=24 are x=2,x=-1+i(sqrt(7))/2 ,x=-1-i(sqrt(7))/2
Find the zeros of 4x^3-5x+2=24
The zeros of 4x^3-5x+2=24 are x=2,x=-1+i(sqrt(7))/2 ,x=-1-i(sqrt(7))/2