Linear Algebra Examples

$2 x (4 - x) = 8$

Step 1

Divide each term in $2 x (4 - x) = 8$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 x (4 - x) = 8$ by $2$ .

$\frac{2 x (4 - x)}{2} = \frac{8}{2}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify terms.

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Step 1.2.1.1

Cancel the common factor of $2$ .

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Step 1.2.1.1.1

Cancel the common factor.

$\frac{2 x (4 - x)}{2} = \frac{8}{2}$

Step 1.2.1.1.2

Divide $x (4 - x)$ by $1$ .

$x (4 - x) = \frac{8}{2}$

Step 1.2.1.2

Apply the distributive property.

$x \cdot 4 + x (- x) = \frac{8}{2}$

Step 1.2.1.3

Reorder.

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Step 1.2.1.3.1

Move $4$ to the left of $x$ .

$4 \cdot x + x (- x) = \frac{8}{2}$

Step 1.2.1.3.2

Rewrite using the commutative property of multiplication.

$4 \cdot x - x \cdot x = \frac{8}{2}$

Step 1.2.2

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

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Step 1.2.2.1

Move $x$ .

$4 x - (x \cdot x) = \frac{8}{2}$

Step 1.2.2.2

Multiply $x$ by $x$ .

$4 x - x^{2} = \frac{8}{2}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $8$ by $2$ .

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

Step 2

Subtract $4$ from both sides of the equation.

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

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $- 1$ out of $4 x - x^{2} - 4$ .

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Step 3.1.1

Reorder $4 x$ and $- x^{2}$ .

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

Step 3.1.2

Factor $- 1$ out of $- x^{2}$ .

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

Step 3.1.3

Factor $- 1$ out of $4 x$ .

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

Step 3.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$- (x^{2}) - (- 4 x) - 1 \cdot 4 = 0$

Step 3.1.5

Factor $- 1$ out of $- (x^{2}) - (- 4 x)$ .

$- (x^{2} - 4 x) - 1 \cdot 4 = 0$

Step 3.1.6

Factor $- 1$ out of $- (x^{2} - 4 x) - 1 (4)$ .

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

Step 3.2

Factor using the perfect square rule.

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Step 3.2.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.2.2

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

$4 x = 2 \cdot x \cdot 2$

Step 3.2.3

Rewrite the polynomial.

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

Step 3.2.4

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

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

Step 4

Divide each term in $- {(x - 2)}^{2} = 0$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- {(x - 2)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 2)}^{2}}{- 1} = \frac{0}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{{(x - 2)}^{2}}{1} = \frac{0}{- 1}$

Step 4.2.2

Divide ${(x - 2)}^{2}$ by $1$ .

${(x - 2)}^{2} = \frac{0}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 1$ .

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

Step 5

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

$x - 2 = 0$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

The domain is the set of all valid $x$ values.

${x | x = 2}$

Step 8