Calculus Examples

$3 x^{2} + 6 x - 9 = 0$

Step 1

Factor $3$ out of $3 x^{2} + 6 x - 9$ .

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

$3 (x^{2}) + 6 x - 9 = 0$

Factor $3$ out of $6 x$ .

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

Factor $3$ out of $- 9$ .

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

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

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

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

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

Step 2

Factor.

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Factor $x^{2} + 2 x - 3$ using the AC method.

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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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$3 ((x - 1) (x + 3)) = 0$

Remove unnecessary parentheses.

$3 (x - 1) (x + 3) = 0$

Step 3

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

$x - 1 = 0$

$x + 3 = 0$

Step 4

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

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

$x = 1, - 3$