(2-c)(4+2c+c^2)=0

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Solution for (2-c)(4+2c+c^2)=0 equation:


Simplifying
(2 + -1c)(4 + 2c + c2) = 0

Multiply (2 + -1c) * (4 + 2c + c2)
(2(4 + 2c + c2) + -1c * (4 + 2c + c2)) = 0
((4 * 2 + 2c * 2 + c2 * 2) + -1c * (4 + 2c + c2)) = 0
((8 + 4c + 2c2) + -1c * (4 + 2c + c2)) = 0
(8 + 4c + 2c2 + (4 * -1c + 2c * -1c + c2 * -1c)) = 0
(8 + 4c + 2c2 + (-4c + -2c2 + -1c3)) = 0

Reorder the terms:
(8 + 4c + -4c + 2c2 + -2c2 + -1c3) = 0

Combine like terms: 4c + -4c = 0
(8 + 0 + 2c2 + -2c2 + -1c3) = 0
(8 + 2c2 + -2c2 + -1c3) = 0

Combine like terms: 2c2 + -2c2 = 0
(8 + 0 + -1c3) = 0
(8 + -1c3) = 0

Solving
8 + -1c3 = 0

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '-8' to each side of the equation.
8 + -8 + -1c3 = 0 + -8

Combine like terms: 8 + -8 = 0
0 + -1c3 = 0 + -8
-1c3 = 0 + -8

Combine like terms: 0 + -8 = -8
-1c3 = -8

Divide each side by '-1'.
c3 = 8

Simplifying
c3 = 8

Reorder the terms:
-8 + c3 = 8 + -8

Combine like terms: 8 + -8 = 0
-8 + c3 = 0

The solution to this equation could not be determined.

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