Solve for a 15(2a-2)=5(a^2-1)

$15 (2 a - 2) = 5 (a^{2} - 1)$

Simplify $15 (2 a - 2)$ .

Start simplifying.

$0 + 0 + 15 (2 a - 2) = 5 (a^{2} - 1)$

Simplify by adding zeros.

$15 (2 a - 2) = 5 (a^{2} - 1)$

Apply the distributive property.

$15 (2 a) + 15 \cdot - 2 = 5 (a^{2} - 1)$

Multiply.

Multiply $2$ by $15$ .

$30 a + 15 \cdot - 2 = 5 (a^{2} - 1)$

Multiply $15$ by $- 2$ .

$30 a - 30 = 5 (a^{2} - 1)$

Simplify $5 (a^{2} - 1)$ .

Apply the distributive property.

$30 a - 30 = 5 a^{2} + 5 \cdot - 1$

Multiply $5$ by $- 1$ .

$30 a - 30 = 5 a^{2} - 5$

Subtract $5 a^{2}$ from both sides of the equation.

$30 a - 30 - 5 a^{2} = - 5$

Add $5$ to both sides of the equation.

$30 a - 30 - 5 a^{2} + 5 = 0$

Add $- 30$ and $5$ .

$30 a - 5 a^{2} - 25 = 0$

Factor the left side of the equation.

Factor $- 5$ out of $30 a - 5 a^{2} - 25$ .

Reorder $30 a$ and $- 5 a^{2}$ .

$- 5 a^{2} + 30 a - 25 = 0$

Factor $- 5$ out of $- 5 a^{2}$ .

$- 5 a^{2} + 30 a - 25 = 0$

Factor $- 5$ out of $30 a$ .

$- 5 a^{2} - 5 (- 6 a) - 25 = 0$

Factor $- 5$ out of $- 25$ .

$- 5 a^{2} - 5 (- 6 a) - 5 \cdot 5 = 0$

Factor $- 5$ out of $- 5 (a^{2}) - 5 (- 6 a)$ .

$- 5 (a^{2} - 6 a) - 5 \cdot 5 = 0$

Factor $- 5$ out of $- 5 (a^{2} - 6 a) - 5 (5)$ .

$- 5 (a^{2} - 6 a + 5) = 0$

Factor.

Factor $a^{2} - 6 a + 5$ 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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$- 5 ((a - 5) (a - 1)) = 0$

Remove unnecessary parentheses.

$- 5 (a - 5) (a - 1) = 0$

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

$a - 5 = 0$

$a - 1 = 0$

Set $a - 5$ equal to $0$ and solve for $a$ .

Set $a - 5$ equal to $0$ .

$a - 5 = 0$

Add $5$ to both sides of the equation.

$a = 5$

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

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

$a - 1 = 0$

Add $1$ to both sides of the equation.

$a = 1$

The final solution is all the values that make $- 5 (a - 5) (a - 1) = 0$ true.

$a = 5, 1$

Solve for a 15(2a-2)=5(a^2-1)

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