Precalculus Examples

$f (x) = x^{4} - x^{3} - 7 x^{2} + 5 x + 10$

Step 1

Set $x^{4} - x^{3} - 7 x^{2} + 5 x + 10$ equal to $0$ .

$x^{4} - x^{3} - 7 x^{2} + 5 x + 10 = 0$

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.

$- x^{3} + 5 x + x^{4} - 7 x^{2} + 10 = 0$

Step 2.1.2

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

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Step 2.1.2.1

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

$- x \cdot x^{2} + 5 x + x^{4} - 7 x^{2} + 10 = 0$

Step 2.1.2.2

Factor $- x$ out of $5 x$ .

$- x \cdot x^{2} - x \cdot - 5 + x^{4} - 7 x^{2} + 10 = 0$

Step 2.1.2.3

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

$- x (x^{2} - 5) + x^{4} - 7 x^{2} + 10 = 0$

Step 2.1.3

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

$- x (x^{2} - 5) + {(x^{2})}^{2} - 7 x^{2} + 10 = 0$

Step 2.1.4

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

$- x (x^{2} - 5) + u^{2} - 7 u + 10 = 0$

Step 2.1.5

Factor $u^{2} - 7 u + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Step 2.1.5.2

Write the factored form using these integers.

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

Step 2.1.6

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

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

Step 2.1.7

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

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Step 2.1.7.1

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

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

Step 2.1.7.2

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

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

Step 2.1.8

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

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

Step 2.1.9

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

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Step 2.1.9.1

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 $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 2.1.9.2

Write the factored form using these integers.

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

Step 2.1.10

Factor.

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Step 2.1.10.1

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

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

Step 2.1.10.2

Remove unnecessary parentheses.

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

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 $0$ .

$x^{2} - 5 = 0$

$x - 2 = 0$

$x + 1 = 0$

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

Add $5$ to 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

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.

$x = \sqrt{5}$

Step 2.3.2.3.2

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

$x = - \sqrt{5}$

Step 2.3.2.3.3

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

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

Step 2.4

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

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Step 2.4.1

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

$x - 2 = 0$

Step 2.4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.5

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

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Step 2.5.1

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

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

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

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 2.23606797 \dots, - 2.23606797 \dots, 2, - 1$

Step 4