Algebra Examples

$f (x) = 36 + 36 x - 13 x^{2} - 13 x^{3} + x^{4} + x^{5}$

Step 1

Set $36 + 36 x - 13 x^{2} - 13 x^{3} + x^{4} + x^{5}$ equal to $0$ .

$36 + 36 x - 13 x^{2} - 13 x^{3} + x^{4} + x^{5} = 0$

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.

$36 - 13 x^{2} + x^{4} + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.2

Rewrite the middle term.

$36 + 2 \cdot (6 x^{2}) - 25 x^{2} + x^{4} + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.3

Rearrange terms.

$36 + 2 \cdot (6 x^{2}) + x^{4} - 25 x^{2} + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.4

Factor first three terms by perfect square rule.

${(6 + x^{2})}^{2} - 25 x^{2} + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.5

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

${(6 + x^{2})}^{2} - {(5 x)}^{2} + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6 + x^{2}$ and $b = 5 x$ .

$(6 + x^{2} + 5 x) (6 + x^{2} - (5 x)) + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.7

Simplify.

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Step 2.1.7.1

Factor $6 + x^{2} + 5 x$ using the AC method.

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Step 2.1.7.1.1

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 $6$ and whose sum is $5$ .

$2, 3$

Step 2.1.7.1.2

Write the factored form using these integers.

$(x + 2) (x + 3) (6 + x^{2} - (5 x)) + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.7.2

Multiply $5$ by $- 1$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + 36 x - 13 x^{3} + x^{5} = 0$

Step 2.1.8

Factor $x$ out of $36 x - 13 x^{3} + x^{5}$ .

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Step 2.1.8.1

Factor $x$ out of $36 x$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x \cdot 36 - 13 x^{3} + x^{5} = 0$

Step 2.1.8.2

Factor $x$ out of $- 13 x^{3}$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x \cdot 36 + x (- 13 x^{2}) + x^{5} = 0$

Step 2.1.8.3

Factor $x$ out of $x^{5}$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x \cdot 36 + x (- 13 x^{2}) + x \cdot x^{4} = 0$

Step 2.1.8.4

Factor $x$ out of $x \cdot 36 + x (- 13 x^{2})$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x \cdot (36 - 13 x^{2}) + x \cdot x^{4} = 0$

Step 2.1.8.5

Factor $x$ out of $x \cdot (36 - 13 x^{2}) + x \cdot x^{4}$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x (36 - 13 x^{2} + x^{4}) = 0$

Step 2.1.9

Rewrite the middle term.

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x (36 + 2 \cdot (6 x^{2}) - 25 x^{2} + x^{4}) = 0$

Step 2.1.10

Rearrange terms.

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x (36 + 2 \cdot (6 x^{2}) + x^{4} - 25 x^{2}) = 0$

Step 2.1.11

Factor first three terms by perfect square rule.

$(x + 2) (x + 3) (6 + x^{2} - 5 x) + x ({(6 + x^{2})}^{2} - 25 x^{2}) = 0$

Step 2.1.12

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

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

Step 2.1.13

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 6 + x^{2}$ and $b = 5 x$ .

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

Step 2.1.14

Factor.

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Step 2.1.14.1

Simplify.

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Step 2.1.14.1.1

Factor $6 + x^{2} + 5 x$ using the AC method.

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Step 2.1.14.1.1.1

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 $6$ and whose sum is $5$ .

$2, 3$

Step 2.1.14.1.1.2

Write the factored form using these integers.

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

Step 2.1.14.1.2

Multiply $5$ by $- 1$ .

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

Step 2.1.14.2

Remove unnecessary parentheses.

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

Step 2.1.15

Factor $(x + 2) (x + 3) (6 + x^{2} - 5 x)$ out of $(x + 2) (x + 3) (6 + x^{2} - 5 x) + x (x + 2) (x + 3) (6 + x^{2} - 5 x)$ .

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Step 2.1.15.1

Factor $(x + 2) (x + 3) (6 + x^{2} - 5 x)$ out of $(x + 2) (x + 3) (6 + x^{2} - 5 x)$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) (1) + x (x + 2) (x + 3) (6 + x^{2} - 5 x) = 0$

Step 2.1.15.2

Factor $(x + 2) (x + 3) (6 + x^{2} - 5 x)$ out of $x (x + 2) (x + 3) (6 + x^{2} - 5 x)$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) (1) + (x + 2) (x + 3) (6 + x^{2} - 5 x) (x) = 0$

Step 2.1.15.3

Factor $(x + 2) (x + 3) (6 + x^{2} - 5 x)$ out of $(x + 2) (x + 3) (6 + x^{2} - 5 x) (1) + (x + 2) (x + 3) (6 + x^{2} - 5 x) (x)$ .

$(x + 2) (x + 3) (6 + x^{2} - 5 x) (1 + x) = 0$

Step 2.1.16

Factor.

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Step 2.1.16.1

Factor $6 + x^{2} - 5 x$ using the AC method.

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Step 2.1.16.1.1

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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 2.1.16.1.2

Write the factored form using these integers.

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

Step 2.1.16.2

Remove unnecessary parentheses.

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

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 $0$ .

$x + 2 = 0$

$x + 3 = 0$

$x - 3 = 0$

$x - 2 = 0$

$1 + x = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 2 = 0$

Step 2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.4

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

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Step 2.4.1

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

$x + 3 = 0$

Step 2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.5

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

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Step 2.5.1

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

$x - 3 = 0$

Step 2.5.2

Add $3$ to both sides of the equation.

$x = 3$

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

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7

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

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Step 2.7.1

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

$1 + x = 0$

Step 2.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.8

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

$x = - 2, - 3, 3, 2, - 1$

Step 3