Precalculus Examples

$f (x) = x^{4} - 4 x^{3} - 10 x^{2} - 20 x - 75$

Step 1

Set $x^{4} - 4 x^{3} - 10 x^{2} - 20 x - 75$ equal to $0$ .

$x^{4} - 4 x^{3} - 10 x^{2} - 20 x - 75 = 0$

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.

$- 4 x^{3} - 20 x + x^{4} - 10 x^{2} - 75 = 0$

Step 2.1.2

Factor $- 4 x$ out of $- 4 x^{3} - 20 x$ .

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Step 2.1.2.1

Factor $- 4 x$ out of $- 4 x^{3}$ .

$- 4 x \cdot x^{2} - 20 x + x^{4} - 10 x^{2} - 75 = 0$

Step 2.1.2.2

Factor $- 4 x$ out of $- 20 x$ .

$- 4 x \cdot x^{2} - 4 x \cdot 5 + x^{4} - 10 x^{2} - 75 = 0$

Step 2.1.2.3

Factor $- 4 x$ out of $- 4 x (x^{2}) - 4 x (5)$ .

$- 4 x (x^{2} + 5) + x^{4} - 10 x^{2} - 75 = 0$

Step 2.1.3

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

$- 4 x (x^{2} + 5) + {(x^{2})}^{2} - 10 x^{2} - 75 = 0$

Step 2.1.4

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

$- 4 x (x^{2} + 5) + u^{2} - 10 u - 75 = 0$

Step 2.1.5

Factor $u^{2} - 10 u - 75$ using the AC method.

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Step 2.1.5.1

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 $- 75$ and whose sum is $- 10$ .

$- 15, 5$

Step 2.1.5.2

Write the factored form using these integers.

$- 4 x (x^{2} + 5) + (u - 15) (u + 5) = 0$

Step 2.1.6

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

$- 4 x (x^{2} + 5) + (x^{2} - 15) (x^{2} + 5) = 0$

Step 2.1.7

Factor $x^{2} + 5$ out of $- 4 x (x^{2} + 5) + (x^{2} - 15) (x^{2} + 5)$ .

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Step 2.1.7.1

Factor $x^{2} + 5$ out of $- 4 x (x^{2} + 5)$ .

$(x^{2} + 5) (- 4 x) + (x^{2} - 15) (x^{2} + 5) = 0$

Step 2.1.7.2

Factor $x^{2} + 5$ out of $(x^{2} - 15) (x^{2} + 5)$ .

$(x^{2} + 5) (- 4 x) + (x^{2} + 5) (x^{2} - 15) = 0$

Step 2.1.7.3

Factor $x^{2} + 5$ out of $(x^{2} + 5) (- 4 x) + (x^{2} + 5) (x^{2} - 15)$ .

$(x^{2} + 5) (- 4 x + x^{2} - 15) = 0$

Step 2.1.8

Reorder terms.

$(x^{2} + 5) (x^{2} - 4 x - 15) = 0$

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 $0$ .

$x^{2} + 5 = 0$

$x^{2} - 4 x - 15 = 0$

Step 2.3

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

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Step 2.3.1

Set $x^{2} + 5$ equal to $0$ .

$x^{2} + 5 = 0$

Step 2.3.2

Solve $x^{2} + 5 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

Step 2.3.2.2

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

$x = \pm \sqrt{- 5}$

Step 2.3.2.3

Simplify $\pm \sqrt{- 5}$ .

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Step 2.3.2.3.1

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Step 2.3.2.3.2

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

Step 2.3.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{5}$

Step 2.3.2.4

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

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Step 2.3.2.4.1

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

$x = i \sqrt{5}$

Step 2.3.2.4.2

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

$x = - i \sqrt{5}$

Step 2.3.2.4.3

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 2.4

Set $x^{2} - 4 x - 15$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} - 4 x - 15$ equal to $0$ .

$x^{2} - 4 x - 15 = 0$

Step 2.4.2

Solve $x^{2} - 4 x - 15 = 0$ for $x$ .

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Step 2.4.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.4.2.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Step 2.4.2.3

Simplify.

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Step 2.4.2.3.1

Simplify the numerator.

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Step 2.4.2.3.1.1

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Step 2.4.2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 15$ .

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Step 2.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 15}}{2 \cdot 1}$

Step 2.4.2.3.1.2.2

Multiply $- 4$ by $- 15$ .

$x = \frac{4 \pm \sqrt{16 + 60}}{2 \cdot 1}$

Step 2.4.2.3.1.3

Add $16$ and $60$ .

$x = \frac{4 \pm \sqrt{76}}{2 \cdot 1}$

Step 2.4.2.3.1.4

Rewrite $76$ as $2^{2} \cdot 19$ .

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Step 2.4.2.3.1.4.1

Factor $4$ out of $76$ .

$x = \frac{4 \pm \sqrt{4 (19)}}{2 \cdot 1}$

Step 2.4.2.3.1.4.2

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 19}}{2 \cdot 1}$

Step 2.4.2.3.1.5

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{19}}{2 \cdot 1}$

Step 2.4.2.3.2

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{19}}{2}$

Step 2.4.2.3.3

Simplify $\frac{4 \pm 2 \sqrt{19}}{2}$ .

$x = 2 \pm \sqrt{19}$

Step 2.4.2.4

The final answer is the combination of both solutions.

$x = 2 + \sqrt{19}, 2 - \sqrt{19}$

Step 2.5

The final solution is all the values that make $(x^{2} + 5) (x^{2} - 4 x - 15) = 0$ true.

$x = i \sqrt{5}, - i \sqrt{5}, 2 + \sqrt{19}, 2 - \sqrt{19}$

Step 3