Pre-Algebra Examples

$2 x (2 x - 10) = - 30 + 6 x$

Step 1

Simplify $2 x (2 x - 10)$ .

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Step 1.1

Rewrite.

$0 + 0 + 2 x (2 x - 10) = - 30 + 6 x$

Step 1.2

Simplify by multiplying through.

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Step 1.2.1

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 10 = - 30 + 6 x$

Step 1.2.2

Simplify the expression.

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Step 1.2.2.1

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot - 10 = - 30 + 6 x$

Step 1.2.2.2

Multiply $- 10$ by $2$ .

$2 \cdot 2 x \cdot x - 20 x = - 30 + 6 x$

Step 1.3

Simplify each term.

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Step 1.3.1

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

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Step 1.3.1.1

Move $x$ .

$2 \cdot 2 (x \cdot x) - 20 x = - 30 + 6 x$

Step 1.3.1.2

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} - 20 x = - 30 + 6 x$

Step 1.3.2

Multiply $2$ by $2$ .

$4 x^{2} - 20 x = - 30 + 6 x$

Step 2

Move all terms containing $x$ to the left side of the equation.

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Step 2.1

Subtract $6 x$ from both sides of the equation.

$4 x^{2} - 20 x - 6 x = - 30$

Step 2.2

Subtract $6 x$ from $- 20 x$ .

$4 x^{2} - 26 x = - 30$

Step 3

Add $30$ to both sides of the equation.

$4 x^{2} - 26 x + 30 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $2$ out of $4 x^{2} - 26 x + 30$ .

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Step 4.1.1

Factor $2$ out of $4 x^{2}$ .

$2 (2 x^{2}) - 26 x + 30 = 0$

Step 4.1.2

Factor $2$ out of $- 26 x$ .

$2 (2 x^{2}) + 2 (- 13 x) + 30 = 0$

Step 4.1.3

Factor $2$ out of $30$ .

$2 (2 x^{2}) + 2 (- 13 x) + 2 (15) = 0$

Step 4.1.4

Factor $2$ out of $2 (2 x^{2}) + 2 (- 13 x)$ .

$2 (2 x^{2} - 13 x) + 2 (15) = 0$

Step 4.1.5

Factor $2$ out of $2 (2 x^{2} - 13 x) + 2 (15)$ .

$2 (2 x^{2} - 13 x + 15) = 0$

Step 4.2

Factor.

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Step 4.2.1

Factor by grouping.

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Step 4.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = - 13$ .

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Step 4.2.1.1.1

Factor $- 13$ out of $- 13 x$ .

$2 (2 x^{2} - 13 x + 15) = 0$

Step 4.2.1.1.2

Rewrite $- 13$ as $- 3$ plus $- 10$

$2 (2 x^{2} + (- 3 - 10) x + 15) = 0$

Step 4.2.1.1.3

Apply the distributive property.

$2 (2 x^{2} - 3 x - 10 x + 15) = 0$

Step 4.2.1.2

Factor out the greatest common factor from each group.

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Step 4.2.1.2.1

Group the first two terms and the last two terms.

$2 ((2 x^{2} - 3 x) - 10 x + 15) = 0$

Step 4.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$2 (x (2 x - 3) - 5 (2 x - 3)) = 0$

Step 4.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

$2 ((2 x - 3) (x - 5)) = 0$

Step 4.2.2

Remove unnecessary parentheses.

$2 (2 x - 3) (x - 5) = 0$

Step 5

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

$2 x - 3 = 0$

$x - 5 = 0$

Step 6

Set $2 x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Step 6.2

Solve $2 x - 3 = 0$ for $x$ .

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Step 6.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 6.2.2

Divide each term in $2 x = 3$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 7

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

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Step 7.1

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

$x - 5 = 0$

Step 7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 8

The final solution is all the values that make $2 (2 x - 3) (x - 5) = 0$ true.

$x = \frac{3}{2}, 5$