Solve for y 3y^4-6y^3=-3y^2

$3 y^{4} - 6 y^{3} = - 3 y^{2}$

Add $3 y^{2}$ to both sides of the equation.

$3 y^{4} - 6 y^{3} + 3 y^{2} = 0$

Factor the left side of the equation.

Factor $3 y^{2}$ out of $3 y^{4} - 6 y^{3} + 3 y^{2}$ .

Factor $3 y^{2}$ out of $3 y^{4}$ .

$3 y^{2} (y^{2}) - 6 y^{3} + 3 y^{2} = 0$

Factor $3 y^{2}$ out of $- 6 y^{3}$ .

$3 y^{2} (y^{2}) + 3 y^{2} (- 2 y) + 3 y^{2} = 0$

Factor $3 y^{2}$ out of $3 y^{2}$ .

$3 y^{2} (y^{2}) + 3 y^{2} (- 2 y) + 3 y^{2} (1) = 0$

Factor $3 y^{2}$ out of $3 y^{2} (y^{2}) + 3 y^{2} (- 2 y)$ .

$3 y^{2} (y^{2} - 2 y) + 3 y^{2} (1) = 0$

Factor $3 y^{2}$ out of $3 y^{2} (y^{2} - 2 y) + 3 y^{2} (1)$ .

$3 y^{2} (y^{2} - 2 y + 1) = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$3 y^{2} (y^{2} - 2 y + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 y = 2 \cdot y \cdot 1$

Rewrite the polynomial.

$3 y^{2} (y^{2} - 2 \cdot y \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 1$ .

$3 y^{2} {(y - 1)}^{2} = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y^{2} = 0$

${(y - 1)}^{2} = 0$

Set $y^{2}$ equal to $0$ and solve for $y$ .

Set $y^{2}$ equal to $0$ .

$y^{2} = 0$

Solve $y^{2} = 0$ for $y$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$y = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 0$

Plus or minus $0$ is $0$ .

$y = 0$

Set ${(y - 1)}^{2}$ equal to $0$ and solve for $y$ .

Set ${(y - 1)}^{2}$ equal to $0$ .

${(y - 1)}^{2} = 0$

Solve ${(y - 1)}^{2} = 0$ for $y$ .

Set the $y - 1$ equal to $0$ .

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

The final solution is all the values that make $3 y^{2} {(y - 1)}^{2} = 0$ true.

$y = 0, 1$

Solve for y 3y^4-6y^3=-3y^2

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