What are the solutions to the equation x^2(x-3)=12-4x ?

The solutions to the equation x^2(x-3)=12-4x are x=3,x=2i,x=-2i

Find the zeros of x^2(x-3)=12-4x

The zeros of x^2(x-3)=12-4x are x=3,x=2i,x=-2i

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x^{2} (x - 3) = 12 - 4 x

x = 3, x = 2 i, x = - 2 i

Solution steps

x^{2} (x - 3) = 12 - 4 x

Expand

x^{2} (x - 3) : x^{3} - 3 x^{2}

x^{3} - 3 x^{2} = 12 - 4 x

Move

4 x

to the left side

x^{3} - 3 x^{2} + 4 x = 12

Move

12

to the left side

x^{3} - 3 x^{2} + 4 x - 12 = 0

Factor

x^{3} - 3 x^{2} + 4 x - 12 : (x - 3) (x^{2} + 4)

(x - 3) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

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

The solutions are

x = 3, x = 2 i, x = - 2 i

x^{6} + x^{4} - 2 x^{3} - 2 x + 1 = 0

- λ^{3} - λ^{2} + 8 λ + 12 = 0

6 x^{4} + 8 x^{2} = 26 x^{3}

x^{3} - 8 x^{2} + 217 x + 224 = 0

\frac{12 + x}{2} = (x^{2} + 36)^{2}

What are the solutions to the equation x^2(x-3)=12-4x ?
The solutions to the equation x^2(x-3)=12-4x are x=3,x=2i,x=-2i
Find the zeros of x^2(x-3)=12-4x
The zeros of x^2(x-3)=12-4x are x=3,x=2i,x=-2i