Algebra Examples

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

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

Step 1

Set

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

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

0

$0$ .

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

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

Step 2

Solve for

x

$x$ .

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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

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

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

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

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

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

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

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

Step 2.1.5

Rewrite

25 x^{2}

$25 x^{2}$ as

{(5 x)}^{2}

${(5 x)}^{2}$ .

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

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

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)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = 6 + x^{2}

$a = 6 + x^{2}$ and

b = 5 x

$b = 5 x$ .

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

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

Step 2.1.7

Simplify.

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Step 2.1.7.1

Factor

6 + x^{2} + 5 x

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

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Step 2.1.7.1.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

6

$6$ and whose sum is

5

$5$ .

2, 3

$2, 3$

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

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

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