Pre-Algebra Examples

$70 = (2 x + 4) x$

Step 1

Rewrite the equation as $(2 x + 4) x = 70$ .

$(2 x + 4) x = 70$

Step 2

Simplify $(2 x + 4) x$ .

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Step 2.1

Apply the distributive property.

$2 x \cdot x + 4 x = 70$

Step 2.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.1

Move $x$ .

$2 (x \cdot x) + 4 x = 70$

Step 2.2.2

Multiply $x$ by $x$ .

$2 x^{2} + 4 x = 70$

Step 3

Subtract $70$ from both sides of the equation.

$2 x^{2} + 4 x - 70 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $2$ out of $2 x^{2} + 4 x - 70$ .

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Step 4.1.1

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

$2 (x^{2}) + 4 x - 70 = 0$

Step 4.1.2

Factor $2$ out of $4 x$ .

$2 (x^{2}) + 2 (2 x) - 70 = 0$

Step 4.1.3

Factor $2$ out of $- 70$ .

$2 x^{2} + 2 (2 x) + 2 \cdot - 35 = 0$

Step 4.1.4

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

$2 (x^{2} + 2 x) + 2 \cdot - 35 = 0$

Step 4.1.5

Factor $2$ out of $2 (x^{2} + 2 x) + 2 \cdot - 35$ .

$2 (x^{2} + 2 x - 35) = 0$

Step 4.2

Factor.

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Step 4.2.1

Factor $x^{2} + 2 x - 35$ using the AC method.

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Step 4.2.1.1

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 $- 35$ and whose sum is $2$ .

$- 5, 7$

Step 4.2.1.2

Write the factored form using these integers.

$2 ((x - 5) (x + 7)) = 0$

Step 4.2.2

Remove unnecessary parentheses.

$2 (x - 5) (x + 7) = 0$

Step 5

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

$x - 5 = 0$

$x + 7 = 0$

Step 6

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

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Step 6.1

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

$x - 5 = 0$

Step 6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Set $x + 7$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 7.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8

The final solution is all the values that make $2 (x - 5) (x + 7) = 0$ true.

$x = 5, - 7$