Pre-Algebra Examples

$x (x - 16) = - 32 (x + 2)$

Step 1

Simplify $x (x - 16)$ .

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Step 1.1

Rewrite.

$0 + 0 + x (x - 16) = - 32 (x + 2)$

Step 1.2

Simplify by adding zeros.

$x (x - 16) = - 32 (x + 2)$

Step 1.3

Apply the distributive property.

$x \cdot x + x \cdot - 16 = - 32 (x + 2)$

Step 1.4

Simplify the expression.

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Step 1.4.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 16 = - 32 (x + 2)$

Step 1.4.2

Move $- 16$ to the left of $x$ .

$x^{2} - 16 x = - 32 (x + 2)$

Step 2

Simplify $- 32 (x + 2)$ .

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Step 2.1

Apply the distributive property.

$x^{2} - 16 x = - 32 x - 32 \cdot 2$

Step 2.2

Multiply $- 32$ by $2$ .

$x^{2} - 16 x = - 32 x - 64$

Step 3

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

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Step 3.1

Add $32 x$ to both sides of the equation.

$x^{2} - 16 x + 32 x = - 64$

Step 3.2

Add $- 16 x$ and $32 x$ .

$x^{2} + 16 x = - 64$

Step 4

Add $64$ to both sides of the equation.

$x^{2} + 16 x + 64 = 0$

Step 5

Factor using the perfect square rule.

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Step 5.1

Rewrite $64$ as $8^{2}$ .

$x^{2} + 16 x + 8^{2} = 0$

Step 5.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Step 5.3

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 8 + 8^{2} = 0$

Step 5.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 8$ .

${(x + 8)}^{2} = 0$

Step 6

Set the $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 7

Subtract $8$ from both sides of the equation.

$x = - 8$