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

The solutions to the equation x-4=4x^3-3x^2+2x-4 are x=0,x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

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

The zeros of x-4=4x^3-3x^2+2x-4 are x=0,x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8

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

x = 0, x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

Solution steps

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

Switch sides

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

Move

4

to the right side

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

Move

x

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

4 x^{2} - 3 x + 1 = 0 : x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

The solutions are

x = 0, x = \frac{3}{8} + i \frac{7}{8}, x = \frac{3}{8} - i \frac{7}{8}

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

\frac{[ ( 13 + x ) ( 13 - x ) ( 3 + x ) ( 3 - x ) ]}{16} = 144

x^{5} - 3 x^{2} + x + 10 = 0

(x^{2} + x - 6)^{2} - (x^{2} + 2 x + 1)^{2} = 0

x (x - 1)^{3} (x - 2)^{5} = 0

What are the solutions to the equation x-4=4x^3-3x^2+2x-4 ?
The solutions to the equation x-4=4x^3-3x^2+2x-4 are x=0,x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8
Find the zeros of x-4=4x^3-3x^2+2x-4
The zeros of x-4=4x^3-3x^2+2x-4 are x=0,x= 3/8+i(sqrt(7))/8 ,x= 3/8-i(sqrt(7))/8