Precalculus Examples

$4 x^{4} - 28 x^{3} + 47 x^{2} + 7 x - 12$

Step 1

Set $4 x^{4} - 28 x^{3} + 47 x^{2} + 7 x - 12$ equal to $0$ .

$4 x^{4} - 28 x^{3} + 47 x^{2} + 7 x - 12 = 0$

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

Regroup terms.

$- 28 x^{3} + 7 x + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.2

Factor $- 7 x$ out of $- 28 x^{3} + 7 x$ .

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Step 2.1.2.1

Factor $- 7 x$ out of $- 28 x^{3}$ .

$- 7 x (4 x^{2}) + 7 x + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.2.2

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

$- 7 x (4 x^{2}) - 7 x \cdot - 1 + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.2.3

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

$- 7 x (4 x^{2} - 1) + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.3

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$- 7 x ({(2 x)}^{2} - 1) + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.4

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

$- 7 x ({(2 x)}^{2} - 1^{2}) + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.5

Factor.

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Step 2.1.5.1

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

$- 7 x ((2 x + 1) (2 x - 1)) + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$- 7 x (2 x + 1) (2 x - 1) + 4 x^{4} + 47 x^{2} - 12 = 0$

Step 2.1.6

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 7 x (2 x + 1) (2 x - 1) + 4 {(x^{2})}^{2} + 47 x^{2} - 12 = 0$

Step 2.1.7

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 7 x (2 x + 1) (2 x - 1) + 4 u^{2} + 47 u - 12 = 0$

Step 2.1.8

Factor by grouping.

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Step 2.1.8.1

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 $a \cdot c = 4 \cdot - 12 = - 48$ and whose sum is $b = 47$ .

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Step 2.1.8.1.1

Factor $47$ out of $47 u$ .

$- 7 x (2 x + 1) (2 x - 1) + 4 u^{2} + 47 (u) - 12 = 0$

Step 2.1.8.1.2

Rewrite $47$ as $- 1$ plus $48$

$- 7 x (2 x + 1) (2 x - 1) + 4 u^{2} + (- 1 + 48) u - 12 = 0$

Step 2.1.8.1.3

Apply the distributive property.

$- 7 x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u + 48 u - 12 = 0$

Step 2.1.8.2

Factor out the greatest common factor from each group.

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Step 2.1.8.2.1

Group the first two terms and the last two terms.

$- 7 x (2 x + 1) (2 x - 1) + 4 u^{2} - 1 u + 48 u - 12 = 0$

Step 2.1.8.2.2

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

$- 7 x (2 x + 1) (2 x - 1) + u (4 u - 1) + 12 (4 u - 1) = 0$

Step 2.1.8.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$- 7 x (2 x + 1) (2 x - 1) + (4 u - 1) (u + 12) = 0$

Step 2.1.9

Replace all occurrences of $u$ with $x^{2}$ .

$- 7 x (2 x + 1) (2 x - 1) + (4 x^{2} - 1) (x^{2} + 12) = 0$

Step 2.1.10

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$- 7 x (2 x + 1) (2 x - 1) + ({(2 x)}^{2} - 1) (x^{2} + 12) = 0$

Step 2.1.11

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

$- 7 x (2 x + 1) (2 x - 1) + ({(2 x)}^{2} - 1^{2}) (x^{2} + 12) = 0$

Step 2.1.12

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

$- 7 x (2 x + 1) (2 x - 1) + (2 x + 1) (2 x - 1) (x^{2} + 12) = 0$

Step 2.1.13

Factor $(2 x + 1) (2 x - 1)$ out of $- 7 x (2 x + 1) (2 x - 1) + (2 x + 1) (2 x - 1) (x^{2} + 12)$ .

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Step 2.1.13.1

Factor $(2 x + 1) (2 x - 1)$ out of $- 7 x (2 x + 1) (2 x - 1)$ .

$(2 x + 1) (2 x - 1) (- 7 x) + (2 x + 1) (2 x - 1) (x^{2} + 12) = 0$

Step 2.1.13.2

Factor $(2 x + 1) (2 x - 1)$ out of $(2 x + 1) (2 x - 1) (- 7 x) + (2 x + 1) (2 x - 1) (x^{2} + 12)$ .

$(2 x + 1) (2 x - 1) (- 7 x + x^{2} + 12) = 0$

Step 2.1.14

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

$(2 x + 1) (2 x - 1) (- 7 u + u^{2} + 12) = 0$

Step 2.1.15

Factor $- 7 u + u^{2} + 12$ using the AC method.

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Step 2.1.15.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Step 2.1.15.2

Write the factored form using these integers.

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

Step 2.1.16

Factor.

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Step 2.1.16.1

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

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

Step 2.1.16.2

Remove unnecessary parentheses.

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

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 $0$ .

$2 x + 1 = 0$

$2 x - 1 = 0$

$x - 4 = 0$

$x - 3 = 0$

Step 2.3

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 2.3.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 2.3.2.2

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

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

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

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

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

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

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Step 2.6.1

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

$x - 3 = 0$

Step 2.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.7

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

$x = - \frac{1}{2}, \frac{1}{2}, 4, 3$

Step 3