Algebra Examples

$F (X) = x^{4} - 3 x^{3} - 11 x^{2} + 3 x + 10$

Step 1

Set $x^{4} - 3 x^{3} - 11 x^{2} + 3 x + 10$ equal to $0$ .

$x^{4} - 3 x^{3} - 11 x^{2} + 3 x + 10 = 0$

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.

$- 3 x^{3} + 3 x + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.2

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

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Step 2.1.2.1

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

$- 3 x \cdot x^{2} + 3 x + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.2.2

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

$- 3 x \cdot x^{2} - 3 x \cdot - 1 + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.2.3

Factor $- 3 x$ out of $- 3 x (x^{2}) - 3 x (- 1)$ .

$- 3 x (x^{2} - 1) + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.3

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

$- 3 x (x^{2} - 1^{2}) + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.4

Factor.

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Step 2.1.4.1

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

$- 3 x ((x + 1) (x - 1)) + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$- 3 x (x + 1) (x - 1) + x^{4} - 11 x^{2} + 10 = 0$

Step 2.1.5

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

$- 3 x (x + 1) (x - 1) + {(x^{2})}^{2} - 11 x^{2} + 10 = 0$

Step 2.1.6

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

$- 3 x (x + 1) (x - 1) + u^{2} - 11 u + 10 = 0$

Step 2.1.7

Factor $u^{2} - 11 u + 10$ using the AC method.

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Step 2.1.7.1

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 $- 11$ .

$- 10, - 1$

Step 2.1.7.2

Write the factored form using these integers.

$- 3 x (x + 1) (x - 1) + (u - 10) (u - 1) = 0$

Step 2.1.8

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

$- 3 x (x + 1) (x - 1) + (x^{2} - 10) (x^{2} - 1) = 0$

Step 2.1.9

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

$- 3 x (x + 1) (x - 1) + (x^{2} - 10) (x^{2} - 1^{2}) = 0$

Step 2.1.10

Factor.

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Step 2.1.10.1

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

$- 3 x (x + 1) (x - 1) + (x^{2} - 10) ((x + 1) (x - 1)) = 0$

Step 2.1.10.2

Remove unnecessary parentheses.

$- 3 x (x + 1) (x - 1) + (x^{2} - 10) (x + 1) (x - 1) = 0$

Step 2.1.11

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1) + (x^{2} - 10) (x + 1) (x - 1)$ .

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Step 2.1.11.1

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (- 3 x) + (x^{2} - 10) (x + 1) (x - 1) = 0$

Step 2.1.11.2

Factor $(x + 1) (x - 1)$ out of $(x^{2} - 10) (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) (x^{2} - 10) = 0$

Step 2.1.11.3

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) (x^{2} - 10)$ .

$(x + 1) (x - 1) (- 3 x + x^{2} - 10) = 0$

Step 2.1.12

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

$(x + 1) (x - 1) (- 3 u + u^{2} - 10) = 0$

Step 2.1.13

Factor $- 3 u + u^{2} - 10$ using the AC method.

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Step 2.1.13.1

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 $- 3$ .

$- 5, 2$

Step 2.1.13.2

Write the factored form using these integers.

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

Step 2.1.14

Factor.

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Step 2.1.14.1

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

$(x + 1) (x - 1) ((x - 5) (x + 2)) = 0$

Step 2.1.14.2

Remove unnecessary parentheses.

$(x + 1) (x - 1) (x - 5) (x + 2) = 0$

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 $0$ .

$x + 1 = 0$

$x - 1 = 0$

$x - 5 = 0$

$x + 2 = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 1 = 0$

Step 2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

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 $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.6

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

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Step 2.6.1

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

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

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

$x = - 1, 1, 5, - 2$

Step 3