What are the solutions to the equation x^4+2x^3-8x=16 ?

The solutions to the equation x^4+2x^3-8x=16 are x=-2,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i

Find the zeros of x^4+2x^3-8x=16

The zeros of x^4+2x^3-8x=16 are x=-2,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i

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x^{4} + 2 x^{3} - 8 x = 16

x = - 2, x = 2, x = - 1 + 3 i, x = - 1 - 3 i

Solution steps

x^{4} + 2 x^{3} - 8 x = 16

Move

16

to the left side

x^{4} + 2 x^{3} - 8 x - 16 = 0

Factor

x^{4} + 2 x^{3} - 8 x - 16 : (x + 2) (x - 2) (x^{2} + 2 x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

Solve

x^{2} + 2 x + 4 = 0 : x = - 1 + 3 i, x = - 1 - 3 i

The solutions are

x = - 2, x = 2, x = - 1 + 3 i, x = - 1 - 3 i

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

4 x^{3} - 12 x = 0

60 x^{3} - 30 x = 0

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

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

What are the solutions to the equation x^4+2x^3-8x=16 ?
The solutions to the equation x^4+2x^3-8x=16 are x=-2,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i
Find the zeros of x^4+2x^3-8x=16
The zeros of x^4+2x^3-8x=16 are x=-2,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i