What are the solutions to the equation x^5+8=x^3+8x^2 ?

The solutions to the equation x^5+8=x^3+8x^2 are x=-1,x=1,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i

Find the zeros of x^5+8=x^3+8x^2

The zeros of x^5+8=x^3+8x^2 are x=-1,x=1,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i

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x^{5} + 8 = x^{3} + 8 x^{2}

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

Solution steps

x^{5} + 8 = x^{3} + 8 x^{2}

Move

8 x^{2}

to the left side

x^{5} + 8 - 8 x^{2} = x^{3}

Move

x^{3}

to the left side

x^{5} + 8 - 8 x^{2} - x^{3} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

x^{5} - x^{3} - 8 x^{2} + 8 = 0

Factor

x^{5} - x^{3} - 8 x^{2} + 8 : (x + 1) (x - 1) (x - 2) (x^{2} + 2 x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

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

\frac{1}{16} = x^{4}

z, z^{4} = 4 z^{2} - 8

(1 + x)^{4} = (\frac{59}{42})

y^{3} + 6 y^{2} = 27 y

What are the solutions to the equation x^5+8=x^3+8x^2 ?
The solutions to the equation x^5+8=x^3+8x^2 are x=-1,x=1,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i
Find the zeros of x^5+8=x^3+8x^2
The zeros of x^5+8=x^3+8x^2 are x=-1,x=1,x=2,x=-1+sqrt(3)i,x=-1-sqrt(3)i