Calculus Examples

$8 x - 2 x^{2} - x^{3} = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$8 u - 2 u^{2} - u^{3} = 0$

Step 1.2

Factor $u$ out of $8 u - 2 u^{2} - u^{3}$ .

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Step 1.2.1

Factor $u$ out of $8 u$ .

$u \cdot 8 - 2 u^{2} - u^{3} = 0$

Step 1.2.2

Factor $u$ out of $- 2 u^{2}$ .

$u \cdot 8 + u (- 2 u) - u^{3} = 0$

Step 1.2.3

Factor $u$ out of $- u^{3}$ .

$u \cdot 8 + u (- 2 u) + u (- u^{2}) = 0$

Step 1.2.4

Factor $u$ out of $u \cdot 8 + u (- 2 u)$ .

$u \cdot (8 - 2 u) + u (- u^{2}) = 0$

Step 1.2.5

Factor $u$ out of $u \cdot (8 - 2 u) + u (- u^{2})$ .

$u (8 - 2 u - u^{2}) = 0$

Step 1.3

Rewrite $8$ as $9$ plus $- 1$

$u (9 - 1 - 2 u - u^{2}) = 0$

Step 1.4

Factor using the perfect square rule.

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Step 1.4.1

Rewrite $1$ as $1^{2}$ .

$u (9 - (1^{2} + 2 u + u^{2})) = 0$

Step 1.4.2

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

$2 u = 2 \cdot 1 \cdot u$

Step 1.4.3

Rewrite the polynomial.

$u (9 - (1^{2} + 2 \cdot 1 \cdot u + u^{2})) = 0$

Step 1.4.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 1$ and $b = u$ .

$u (9 - {(1 + u)}^{2}) = 0$

Step 1.5

Rewrite $9$ as $3^{2}$ .

$u (3^{2} - {(1 + u)}^{2}) = 0$

Step 1.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = 1 + u$ .

$u ((3 + 1 + u) (3 - (1 + u))) = 0$

Step 1.7

Factor.

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Step 1.7.1

Simplify.

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Step 1.7.1.1

Remove unnecessary parentheses.

$u ((3 + 1 + u) (3 - (1 + u))) = 0$

Step 1.7.1.2

Add $3$ and $1$ .

$u ((4 + u) (3 - (1 + u))) = 0$

Step 1.7.1.3

Apply the distributive property.

$u ((4 + u) (3 - 1 \cdot 1 - u)) = 0$

Step 1.7.1.4

Multiply $- 1$ by $1$ .

$u ((4 + u) (3 - 1 - u)) = 0$

Step 1.7.1.5

Subtract $1$ from $3$ .

$u ((4 + u) (2 - u)) = 0$

Step 1.7.2

Remove unnecessary parentheses.

$u (4 + u) (2 - u) = 0$

Step 1.8

Replace all occurrences of $u$ with $x$ .

$x (4 + x) (2 - x) = 0$

Step 2

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

$x = 0$

$4 + x = 0$

$2 - x = 0$

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

Set $4 + x$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $4 + x$ equal to $0$ .

$4 + x = 0$

Step 4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

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

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Step 5.1

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

$2 - x = 0$

Step 5.2

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

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Step 5.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 5.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 5.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 5.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 6

The final solution is all the values that make $x (4 + x) (2 - x) = 0$ true.

$x = 0, - 4, 2$