Calculus Examples

$6 x^{2} + 6 x - 72 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $6$ out of $6 x^{2} + 6 x - 72$ .

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Step 1.1.1

Factor $6$ out of $6 x^{2}$ .

$6 (x^{2}) + 6 x - 72 = 0$

Step 1.1.2

Factor $6$ out of $6 x$ .

$6 (x^{2}) + 6 (x) - 72 = 0$

Step 1.1.3

Factor $6$ out of $- 72$ .

$6 (x^{2}) + 6 x + 6 \cdot - 12 = 0$

Step 1.1.4

Factor $6$ out of $6 (x^{2}) + 6 x$ .

$6 (x^{2} + x) + 6 \cdot - 12 = 0$

Step 1.1.5

Factor $6$ out of $6 (x^{2} + x) + 6 \cdot - 12$ .

$6 (x^{2} + x - 12) = 0$

Step 1.2

Factor.

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Step 1.2.1

Factor $x^{2} + x - 12$ using the AC method.

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Step 1.2.1.1

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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 1.2.1.2

Write the factored form using these integers.

$6 ((x - 3) (x + 4)) = 0$

Step 1.2.2

Remove unnecessary parentheses.

$6 (x - 3) (x + 4) = 0$

Step 2

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

$x - 3 = 0$

$x + 4 = 0$

Step 3

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

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Step 3.1

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

$x - 3 = 0$

Step 3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

The final solution is all the values that make $6 (x - 3) (x + 4) = 0$ true.

$x = 3, - 4$