Algebra Examples

(x^{4} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0

$(x^{4} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0$

Step 1

Rewrite

x^{4}

$x^{4}$ as

{(x^{2})}^{2}

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

({(x^{2})}^{2} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0

$({(x^{2})}^{2} + 5 x^{2} - 36) (2 x^{2} + 9 x - 5) = 0$

Step 2

Let

u = x^{2}

$u = x^{2}$ . Substitute

u

$u$ for all occurrences of

x^{2}

$x^{2}$ .

(u^{2} + 5 u - 36) (2 x^{2} + 9 x - 5) = 0

$(u^{2} + 5 u - 36) (2 x^{2} + 9 x - 5) = 0$

Step 3

Factor

u^{2} + 5 u - 36

$u^{2} + 5 u - 36$ using the AC method.

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Step 3.1

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

- 36

$- 36$ and whose sum is

5

$5$ .

- 4, 9

$- 4, 9$

Step 3.2

Write the factored form using these integers.

(u - 4) (u + 9) (2 x^{2} + 9 x - 5) = 0

$(u - 4) (u + 9) (2 x^{2} + 9 x - 5) = 0$

(u - 4) (u + 9) (2 x^{2} + 9 x - 5) = 0

$(u - 4) (u + 9) (2 x^{2} + 9 x - 5) = 0$

Step 4

Replace all occurrences of

u

$u$ with

x^{2}

$x^{2}$ .

$(x^{2} - 4) (x^{2} + 9) (2 x^{2} + 9 x - 5) = 0$

Step 5

Rewrite

$4$ as

$2^{2}$ .

$(x^{2} - 2^{2}) (x^{2} + 9) (2 x^{2} + 9 x - 5) = 0$

Step 6

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

$b = 2$ .

$(x + 2) (x - 2) (x^{2} + 9) (2 x^{2} + 9 x - 5) = 0$

Step 7

Factor.

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Step 7.1

Factor by grouping.

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Step 7.1.1

For a polynomial of the form

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

$a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is

$b = 9$ .

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Step 7.1.1.1

Factor

$9$ out of

$9 x$ .

$(x + 2) (x - 2) (x^{2} + 9) (2 x^{2} + 9 (x) - 5) = 0$

Step 7.1.1.2

Rewrite

$9$ as

$- 1$ plus

$10$

$(x + 2) (x - 2) (x^{2} + 9) (2 x^{2} + (- 1 + 10) x - 5) = 0$

Step 7.1.1.3

Apply the distributive property.

$(x + 2) (x - 2) (x^{2} + 9) (2 x^{2} - 1 x + 10 x - 5) = 0$

Step 7.1.2

Factor out the greatest common factor from each group.

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Step 7.1.2.1

Group the first two terms and the last two terms.

$(x + 2) (x - 2) (x^{2} + 9) ((2 x^{2} - 1 x) + 10 x - 5) = 0$

Step 7.1.2.2

Factor out the greatest common factor (GCF) from each group.

$(x + 2) (x - 2) (x^{2} + 9) (x (2 x - 1) + 5 (2 x - 1)) = 0$

Step 7.1.3

Factor the polynomial by factoring out the greatest common factor,

$2 x - 1$ .

$(x + 2) (x - 2) (x^{2} + 9) ((2 x - 1) (x + 5)) = 0$

Step 7.2

Remove unnecessary parentheses.

$(x + 2) (x - 2) (x^{2} + 9) (2 x - 1) (x + 5) = 0$

Step 8

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x + 2 = 0$

$x - 2 = 0$

$x^{2} + 9 = 0$

$2 x - 1 = 0$

$x + 5 = 0$

Step 9

Set

$x + 2$ equal to

$0$ and solve for

$x$ .

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Step 9.1

Set

$x + 2$ equal to

$0$ .

$x + 2 = 0$

Step 9.2

Subtract

$2$ from both sides of the equation.

$x = - 2$

Step 10

Set

$x - 2$ equal to

$0$ and solve for

$x$ .

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Step 10.1

Set

$x - 2$ equal to

$0$ .

$x - 2 = 0$

Step 10.2

Add

$2$ to both sides of the equation.

$x = 2$

Step 11

Set

$x^{2} + 9$ equal to

$0$ and solve for

$x$ .

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Step 11.1

Set

$x^{2} + 9$ equal to

$0$ .

$x^{2} + 9 = 0$

Step 11.2

Solve

$x^{2} + 9 = 0$ for

$x$ .

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Step 11.2.1

Subtract

$9$ from both sides of the equation.

$x^{2} = - 9$

Step 11.2.2

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

$x = \pm \sqrt{- 9}$

Step 11.2.3

Simplify

$\pm \sqrt{- 9}$ .

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Step 11.2.3.1

Rewrite

$- 9$ as

$- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Step 11.2.3.2

Rewrite

$\sqrt{- 1 (9)}$ as

$\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 11.2.3.3

Rewrite

$\sqrt{- 1}$ as

$i$ .

$x = \pm i \cdot \sqrt{9}$

Step 11.2.3.4

Rewrite

$9$ as

$3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Step 11.2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 3$

Step 11.2.3.6

Move

$3$ to the left of

$i$ .

$x = \pm 3 i$

Step 11.2.4

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

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Step 11.2.4.1

First, use the positive value of the

$\pm$ to find the first solution.

$x = 3 i$

Step 11.2.4.2

Next, use the negative value of the

$\pm$ to find the second solution.

$x = - 3 i$

Step 11.2.4.3

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

$x = 3 i, - 3 i$

Step 12

Set

$2 x - 1$ equal to

$0$ and solve for

$x$ .

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Step 12.1

Set

$2 x - 1$ equal to

$0$ .

$2 x - 1 = 0$

Step 12.2

Solve

$2 x - 1 = 0$ for

$x$ .

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Step 12.2.1

Add

$1$ to both sides of the equation.

$2 x = 1$

Step 12.2.2

Divide each term in

$2 x = 1$ by

$2$ and simplify.

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Step 12.2.2.1

Divide each term in

$2 x = 1$ by

$2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 12.2.2.2

Simplify the left side.

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Step 12.2.2.2.1

Cancel the common factor of

$2$ .

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Step 12.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 12.2.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{1}{2}$

Step 13

Set

$x + 5$ equal to

$0$ and solve for

$x$ .

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Step 13.1

Set

$x + 5$ equal to

$0$ .

$x + 5 = 0$

Step 13.2

Subtract

$5$ from both sides of the equation.

$x = - 5$

Step 14

The final solution is all the values that make

$(x + 2) (x - 2) (x^{2} + 9) (2 x - 1) (x + 5) = 0$ true.

$x = - 2, 2, 3 i, - 3 i, \frac{1}{2}, - 5$