What are the solutions to the equation 1-3(x-4)^3=25 ?

The solutions to the equation 1-3(x-4)^3=25 are x=2,x=5+sqrt(3)i,x=5-sqrt(3)i

Find the zeros of 1-3(x-4)^3=25

The zeros of 1-3(x-4)^3=25 are x=2,x=5+sqrt(3)i,x=5-sqrt(3)i

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1 - 3 (x - 4)^{3} = 25

x = 2, x = 5 + 3 i, x = 5 - 3 i

Solution steps

1 - 3 (x - 4)^{3} = 25

Expand

1 - 3 (x - 4)^{3} : - 3 x^{3} + 36 x^{2} - 144 x + 193

- 3 x^{3} + 36 x^{2} - 144 x + 193 = 25

Move

25

to the left side

- 3 x^{3} + 36 x^{2} - 144 x + 168 = 0

Factor

- 3 x^{3} + 36 x^{2} - 144 x + 168 : - 3 (x - 2) (x^{2} - 10 x + 28)

- 3 (x - 2) (x^{2} - 10 x + 28) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x^{2} - 10 x + 28 = 0

Solve

x - 2 = 0 : x = 2

Solve

x^{2} - 10 x + 28 = 0 : x = 5 + 3 i, x = 5 - 3 i

The solutions are

x = 2, x = 5 + 3 i, x = 5 - 3 i

z^{4} + 4 z^{3} + 8 z^{2} + 80 z + 400 = 0

4 x^{3} = 92

x^{3} - 4 x + 8 = 0

(12 s^{2} + 1)^{2} = 0

x^{2} (x + 2) = 16 (x + 2)

What are the solutions to the equation 1-3(x-4)^3=25 ?
The solutions to the equation 1-3(x-4)^3=25 are x=2,x=5+sqrt(3)i,x=5-sqrt(3)i
Find the zeros of 1-3(x-4)^3=25
The zeros of 1-3(x-4)^3=25 are x=2,x=5+sqrt(3)i,x=5-sqrt(3)i