Solve for w -3(w-1)(-w+2)=0

$- 3 (w - 1) (- w + 2) = 0$

Divide each term in $- 3 (w - 1) (- w + 2) = 0$ by $- 3$ .

$\frac{- 3 (w - 1) (- w + 2)}{- 3} = \frac{0}{- 3}$

Cancel the common factor of $- 3$ .

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Cancel the common factor.

$\frac{- 3 (w - 1) (- w + 2)}{- 3} = \frac{0}{- 3}$

Divide $(w - 1) (- w + 2)$ by $1$ .

$(w - 1) (- w + 2) = \frac{0}{- 3}$

Divide $0$ by $- 3$ .

$(w - 1) (- w + 2) = 0$

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

$w - 1 = 0$

$- w + 2 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$w - 1 = 0$

Add $1$ to both sides of the equation.

$w = 1$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$- w + 2 = 0$

Subtract $2$ from both sides of the equation.

$- w = - 2$

Multiply each term in $- w = - 2$ by $- 1$

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Multiply each term in $- w = - 2$ by $- 1$ .

$(- w) \cdot - 1 = (- 2) \cdot - 1$

Multiply $(- w) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 w = (- 2) \cdot - 1$

Multiply $w$ by $1$ .

$w = (- 2) \cdot - 1$

Multiply $- 2$ by $- 1$ .

$w = 2$

The final solution is all the values that make $\frac{- 3 (w - 1) (- w + 2)}{- 3} = \frac{0}{- 3}$ true.

$w = 1, 2$

Solve for w -3(w-1)(-w+2)=0

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