Algebra Examples

$x^{4} + x^{3} - 5 x^{2} + x - 6$

Step 1

Set $x^{4} + x^{3} - 5 x^{2} + x - 6$ equal to $0$ .

$x^{4} + x^{3} - 5 x^{2} + x - 6 = 0$

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.

$x^{3} + x + x^{4} - 5 x^{2} - 6 = 0$

Step 2.1.2

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

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Step 2.1.2.1

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

$x \cdot x^{2} + x + x^{4} - 5 x^{2} - 6 = 0$

Step 2.1.2.2

Raise $x$ to the power of $1$ .

$x \cdot x^{2} + x + x^{4} - 5 x^{2} - 6 = 0$

Step 2.1.2.3

Factor $x$ out of $x^{1}$ .

$x \cdot x^{2} + x \cdot 1 + x^{4} - 5 x^{2} - 6 = 0$

Step 2.1.2.4

Factor $x$ out of $x \cdot x^{2} + x \cdot 1$ .

$x (x^{2} + 1) + x^{4} - 5 x^{2} - 6 = 0$

Step 2.1.3

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

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

Step 2.1.4

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

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

Step 2.1.5

Factor $u^{2} - 5 u - 6$ 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 $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Step 2.1.5.2

Write the factored form using these integers.

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

Step 2.1.6

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

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

Step 2.1.7

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

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Step 2.1.7.1

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

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

Step 2.1.7.2

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

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

Step 2.1.7.3

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

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

Step 2.1.8

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

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

Step 2.1.9

Factor $u + u^{2} - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.1.9.2

Write the factored form using these integers.

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

Step 2.1.10

Factor.

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Step 2.1.10.1

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

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

Step 2.1.10.2

Remove unnecessary parentheses.

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

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 $0$ .

$x^{2} + 1 = 0$

$x - 2 = 0$

$x + 3 = 0$

Step 2.3

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 2.3.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.3.2.2

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

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

Step 2.3.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 2.3.2.4

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

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Step 2.3.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 2.3.2.4.2

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

$x = - i$

Step 2.3.2.4.3

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

$x = i, - i$

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 $x + 3$ 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

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

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

$x = i, - i, 2, - 3$

Step 3