Solve by Factoring 3z^2+3z-6=0

$3 z^{2} + 3 z - 6 = 0$

Step 1

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

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

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

Factor $3$ out of $3 z$ .

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

Factor $3$ out of $- 6$ .

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

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

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

Factor $3$ out of $3 (z^{2} + z) + 3 \cdot - 2$ .

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

Step 2

Factor.

Factor $z^{2} + z - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

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

$z - 1 = 0$

$z + 2 = 0$

Step 4

Set $z - 1$ equal to $0$ and solve for $z$ .

Set $z - 1$ equal to $0$ .

$z - 1 = 0$

Add $1$ to both sides of the equation.

$z = 1$

Step 5

Set $z + 2$ equal to $0$ and solve for $z$ .

Set $z + 2$ equal to $0$ .

$z + 2 = 0$

Subtract $2$ from both sides of the equation.

$z = - 2$

Step 6

The final solution is all the values that make $3 (z - 1) (z + 2) = 0$ true.

$z = 1, - 2$

Solve by Factoring 3z^2+3z-6=0

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