What are the solutions to the equation -3(x-1)+8(x^3)=6x+7-5x ?

The solutions to the equation -3(x-1)+8(x^3)=6x+7-5x are x=1,x=-1/2+i 1/2 ,x=-1/2-i 1/2

Find the zeros of -3(x-1)+8(x^3)=6x+7-5x

The zeros of -3(x-1)+8(x^3)=6x+7-5x are x=1,x=-1/2+i 1/2 ,x=-1/2-i 1/2

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- 3 (x - 1) + 8 (x^{3}) = 6 x + 7 - 5 x

x = 1, x = - \frac{1}{2} + i \frac{1}{2}, x = - \frac{1}{2} - i \frac{1}{2}

Solution steps

- 3 (x - 1) + 8 (x^{3}) = 6 x + 7 - 5 x

Expand

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

- 3 x + 3 + 8 x^{3} = x + 7

Move

7

to the left side

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

Move

x

to the left side

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

2 x^{2} + 2 x + 1 = 0 : x = - \frac{1}{2} + i \frac{1}{2}, x = - \frac{1}{2} - i \frac{1}{2}

The solutions are

x = 1, x = - \frac{1}{2} + i \frac{1}{2}, x = - \frac{1}{2} - i \frac{1}{2}

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

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

2 y^{3} + y^{2} - 13 y - 6 y = \frac{2}{3}

(x + 1) (- 3 x^{2} + 9 x + 30) = 0

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

What are the solutions to the equation -3(x-1)+8(x^3)=6x+7-5x ?
The solutions to the equation -3(x-1)+8(x^3)=6x+7-5x are x=1,x=-1/2+i 1/2 ,x=-1/2-i 1/2
Find the zeros of -3(x-1)+8(x^3)=6x+7-5x
The zeros of -3(x-1)+8(x^3)=6x+7-5x are x=1,x=-1/2+i 1/2 ,x=-1/2-i 1/2