Calculus Examples

$- 3 x^{3} + 21 x^{2} - 42 x + 24$

Step 1

Set $- 3 x^{3} + 21 x^{2} - 42 x + 24$ equal to $0$ .

$- 3 x^{3} + 21 x^{2} - 42 x + 24 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $3$ out of $- 3 x^{3} + 21 x^{2} - 42 x + 24$ .

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Step 2.1.1.1

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

$3 (- x^{3}) + 21 x^{2} - 42 x + 24 = 0$

Step 2.1.1.2

Factor $3$ out of $21 x^{2}$ .

$3 (- x^{3}) + 3 (7 x^{2}) - 42 x + 24 = 0$

Step 2.1.1.3

Factor $3$ out of $- 42 x$ .

$3 (- x^{3}) + 3 (7 x^{2}) + 3 (- 14 x) + 24 = 0$

Step 2.1.1.4

Factor $3$ out of $24$ .

$3 (- x^{3}) + 3 (7 x^{2}) + 3 (- 14 x) + 3 (8) = 0$

Step 2.1.1.5

Factor $3$ out of $3 (- x^{3}) + 3 (7 x^{2})$ .

$3 (- x^{3} + 7 x^{2}) + 3 (- 14 x) + 3 (8) = 0$

Step 2.1.1.6

Factor $3$ out of $3 (- x^{3} + 7 x^{2}) + 3 (- 14 x)$ .

$3 (- x^{3} + 7 x^{2} - 14 x) + 3 (8) = 0$

Step 2.1.1.7

Factor $3$ out of $3 (- x^{3} + 7 x^{2} - 14 x) + 3 (8)$ .

$3 (- x^{3} + 7 x^{2} - 14 x + 8) = 0$

Step 2.1.2

Regroup terms.

$3 (- x^{3} + 8 + 7 x^{2} - 14 x) = 0$

Step 2.1.3

Factor $- 1$ out of $- x^{3} + 8$ .

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Step 2.1.3.1

Factor $- 1$ out of $- x^{3}$ .

$3 (- (x^{3}) + 8 + 7 x^{2} - 14 x) = 0$

Step 2.1.3.2

Rewrite $8$ as $- 1 (- 8)$ .

$3 (- (x^{3}) - 1 \cdot - 8 + 7 x^{2} - 14 x) = 0$

Step 2.1.3.3

Factor $- 1$ out of $- (x^{3}) - 1 (- 8)$ .

$3 (- (x^{3} - 8) + 7 x^{2} - 14 x) = 0$

Step 2.1.4

Rewrite $8$ as $2^{3}$ .

$3 (- (x^{3} - 2^{3}) + 7 x^{2} - 14 x) = 0$

Step 2.1.5

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

$3 (- ((x - 2) (x^{2} + x \cdot 2 + 2^{2})) + 7 x^{2} - 14 x) = 0$

Step 2.1.6

Factor.

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Step 2.1.6.1

Simplify.

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Step 2.1.6.1.1

Move $2$ to the left of $x$ .

$3 (- ((x - 2) (x^{2} + 2 x + 2^{2})) + 7 x^{2} - 14 x) = 0$

Step 2.1.6.1.2

Raise $2$ to the power of $2$ .

$3 (- ((x - 2) (x^{2} + 2 x + 4)) + 7 x^{2} - 14 x) = 0$

Step 2.1.6.2

Remove unnecessary parentheses.

$3 (- (x - 2) (x^{2} + 2 x + 4) + 7 x^{2} - 14 x) = 0$

Step 2.1.7

Factor $7 x$ out of $7 x^{2} - 14 x$ .

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Step 2.1.7.1

Factor $7 x$ out of $7 x^{2}$ .

$3 (- (x - 2) (x^{2} + 2 x + 4) + 7 x (x) - 14 x) = 0$

Step 2.1.7.2

Factor $7 x$ out of $- 14 x$ .

$3 (- (x - 2) (x^{2} + 2 x + 4) + 7 x (x) + 7 x (- 2)) = 0$

Step 2.1.7.3

Factor $7 x$ out of $7 x (x) + 7 x (- 2)$ .

$3 (- (x - 2) (x^{2} + 2 x + 4) + 7 x (x - 2)) = 0$

Step 2.1.8

Factor $x - 2$ out of $- (x - 2) (x^{2} + 2 x + 4) + 7 x (x - 2)$ .

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Step 2.1.8.1

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

$3 ((x - 2) (- 1 (x^{2} + 2 x + 4)) + 7 x (x - 2)) = 0$

Step 2.1.8.2

Factor $x - 2$ out of $7 x (x - 2)$ .

$3 ((x - 2) (- 1 (x^{2} + 2 x + 4)) + (x - 2) (7 x)) = 0$

Step 2.1.8.3

Factor $x - 2$ out of $(x - 2) (- 1 (x^{2} + 2 x + 4)) + (x - 2) (7 x)$ .

$3 ((x - 2) (- 1 (x^{2} + 2 x + 4) + 7 x)) = 0$

Step 2.1.9

Apply the distributive property.

$3 ((x - 2) (- 1 x^{2} - 1 (2 x) - 1 \cdot 4 + 7 x)) = 0$

Step 2.1.10

Simplify.

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Step 2.1.10.1

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

$3 ((x - 2) (- x^{2} - 1 (2 x) - 1 \cdot 4 + 7 x)) = 0$

Step 2.1.10.2

Multiply $2$ by $- 1$ .

$3 ((x - 2) (- x^{2} - 2 x - 1 \cdot 4 + 7 x)) = 0$

Step 2.1.10.3

Multiply $- 1$ by $4$ .

$3 ((x - 2) (- x^{2} - 2 x - 4 + 7 x)) = 0$

Step 2.1.11

Add $- 2 x$ and $7 x$ .

$3 ((x - 2) (- x^{2} + 5 x - 4)) = 0$

Step 2.1.12

Factor.

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Step 2.1.12.1

Factor.

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Step 2.1.12.1.1

Factor by grouping.

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Step 2.1.12.1.1.1

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 $a \cdot c = - 1 \cdot - 4 = 4$ and whose sum is $b = 5$ .

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Step 2.1.12.1.1.1.1

Factor $5$ out of $5 x$ .

$3 ((x - 2) (- x^{2} + 5 (x) - 4)) = 0$

Step 2.1.12.1.1.1.2

Rewrite $5$ as $1$ plus $4$

$3 ((x - 2) (- x^{2} + (1 + 4) x - 4)) = 0$

Step 2.1.12.1.1.1.3

Apply the distributive property.

$3 ((x - 2) (- x^{2} + 1 x + 4 x - 4)) = 0$

Step 2.1.12.1.1.1.4

Multiply $x$ by $1$ .

$3 ((x - 2) (- x^{2} + x + 4 x - 4)) = 0$

Step 2.1.12.1.1.2

Factor out the greatest common factor from each group.

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Step 2.1.12.1.1.2.1

Group the first two terms and the last two terms.

$3 ((x - 2) ((- x^{2} + x) + 4 x - 4)) = 0$

Step 2.1.12.1.1.2.2

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

$3 ((x - 2) (x (- x + 1) - 4 (- x + 1))) = 0$

Step 2.1.12.1.1.3

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$3 ((x - 2) ((- x + 1) (x - 4))) = 0$

Step 2.1.12.1.2

Remove unnecessary parentheses.

$3 ((x - 2) (- x + 1) (x - 4)) = 0$

Step 2.1.12.2

Remove unnecessary parentheses.

$3 (x - 2) (- x + 1) (x - 4) = 0$

Step 2.2

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

$x - 2 = 0$

$- x + 1 = 0$

$x - 4 = 0$

Step 2.3

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

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Step 2.3.1

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

$x - 2 = 0$

Step 2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.4

Set $- x + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Step 2.4.2

Solve $- x + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.4.2.2.2.2

Divide $x$ by $1$ .

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

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 2.5

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

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Step 2.5.1

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

$x - 4 = 0$

Step 2.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.6

The final solution is all the values that make $3 (x - 2) (- x + 1) (x - 4) = 0$ true.

$x = 2, 1, 4$

Step 3