Precalculus Examples

$f (x) = 2 x^{4} + 3 x^{3} - 11 x^{2} - 9 x + 15$

Step 1

Set $2 x^{4} + 3 x^{3} - 11 x^{2} - 9 x + 15$ equal to $0$ .

$2 x^{4} + 3 x^{3} - 11 x^{2} - 9 x + 15 = 0$

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.

$3 x^{3} - 9 x + 2 x^{4} - 11 x^{2} + 15 = 0$

Step 2.1.2

Factor $3 x$ out of $3 x^{3} - 9 x$ .

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Step 2.1.2.1

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

$3 x (x^{2}) - 9 x + 2 x^{4} - 11 x^{2} + 15 = 0$

Step 2.1.2.2

Factor $3 x$ out of $- 9 x$ .

$3 x (x^{2}) + 3 x (- 3) + 2 x^{4} - 11 x^{2} + 15 = 0$

Step 2.1.2.3

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

$3 x (x^{2} - 3) + 2 x^{4} - 11 x^{2} + 15 = 0$

Step 2.1.3

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

$3 x (x^{2} - 3) + 2 {(x^{2})}^{2} - 11 x^{2} + 15 = 0$

Step 2.1.4

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

$3 x (x^{2} - 3) + 2 u^{2} - 11 u + 15 = 0$

Step 2.1.5

Factor by grouping.

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Step 2.1.5.1

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 $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = - 11$ .

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Step 2.1.5.1.1

Factor $- 11$ out of $- 11 u$ .

$3 x (x^{2} - 3) + 2 u^{2} - 11 u + 15 = 0$

Step 2.1.5.1.2

Rewrite $- 11$ as $- 5$ plus $- 6$

$3 x (x^{2} - 3) + 2 u^{2} + (- 5 - 6) u + 15 = 0$

Step 2.1.5.1.3

Apply the distributive property.

$3 x (x^{2} - 3) + 2 u^{2} - 5 u - 6 u + 15 = 0$

Step 2.1.5.2

Factor out the greatest common factor from each group.

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Step 2.1.5.2.1

Group the first two terms and the last two terms.

$3 x (x^{2} - 3) + 2 u^{2} - 5 u - 6 u + 15 = 0$

Step 2.1.5.2.2

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

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

Step 2.1.5.3

Factor the polynomial by factoring out the greatest common factor, $2 u - 5$ .

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

Step 2.1.6

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

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

Step 2.1.7

Factor $x^{2} - 3$ out of $3 x (x^{2} - 3) + (2 x^{2} - 5) (x^{2} - 3)$ .

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Step 2.1.7.1

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

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

Step 2.1.7.2

Factor $x^{2} - 3$ out of $(2 x^{2} - 5) (x^{2} - 3)$ .

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

Step 2.1.7.3

Factor $x^{2} - 3$ out of $(x^{2} - 3) (3 x) + (x^{2} - 3) (2 x^{2} - 5)$ .

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

Step 2.1.8

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

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

Step 2.1.9

Factor by grouping.

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Step 2.1.9.1

Reorder terms.

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

Step 2.1.9.2

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 3$ .

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Step 2.1.9.2.1

Factor $3$ out of $3 u$ .

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

Step 2.1.9.2.2

Rewrite $3$ as $- 2$ plus $5$

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

Step 2.1.9.2.3

Apply the distributive property.

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

Step 2.1.9.3

Factor out the greatest common factor from each group.

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Step 2.1.9.3.1

Group the first two terms and the last two terms.

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

Step 2.1.9.3.2

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

$(x^{2} - 3) (2 u (u - 1) + 5 (u - 1)) = 0$

Step 2.1.9.4

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

$(x^{2} - 3) ((u - 1) (2 u + 5)) = 0$

Step 2.1.10

Factor.

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Step 2.1.10.1

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

$(x^{2} - 3) ((x - 1) (2 x + 5)) = 0$

Step 2.1.10.2

Remove unnecessary parentheses.

$(x^{2} - 3) (x - 1) (2 x + 5) = 0$

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 $0$ .

$x^{2} - 3 = 0$

$x - 1 = 0$

$2 x + 5 = 0$

Step 2.3

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Step 2.3.2

Solve $x^{2} - 3 = 0$ for $x$ .

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Step 2.3.2.1

Add $3$ to both sides of the equation.

$x^{2} = 3$

Step 2.3.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{3}$

Step 2.3.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.3.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{3}$

Step 2.3.2.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{3}$

Step 2.3.2.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{3}, - \sqrt{3}$

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

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Set $2 x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Step 2.5.2

Solve $2 x + 5 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 2.5.2.2

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 2.6

The final solution is all the values that make $(x^{2} - 3) (x - 1) (2 x + 5) = 0$ true.

$x = \sqrt{3}, - \sqrt{3}, 1, - \frac{5}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{3}, - \sqrt{3}, 1, - \frac{5}{2}$

Decimal Form:

$x = 1.73205080 \dots, - 1.73205080 \dots, 1, - 2.5$

Mixed Number Form:

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

Step 4