Algebra Examples

${(x^{2} + x - 12)}^{5} {(x - 1 + \sqrt{7})}^{3}$

Step 1

Set ${(x^{2} + x - 12)}^{5} {(x - 1 + \sqrt{7})}^{3}$ equal to $0$ .

${(x^{2} + x - 12)}^{5} {(x - 1 + \sqrt{7})}^{3} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor.

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Step 2.1.1

Factor $x^{2} + x - 12$ using the AC method.

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Step 2.1.1.1

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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Step 2.1.1.2

Write the factored form using these integers.

${((x - 3) (x + 4))}^{5} {(x - 1 + \sqrt{7})}^{3} = 0$

Step 2.1.2

Apply the product rule to $(x - 3) (x + 4)$ .

${(x - 3)}^{5} {(x + 4)}^{5} {(x - 1 + \sqrt{7})}^{3} = 0$

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 $0$ .

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

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

${(x - 1 + \sqrt{7})}^{3} = 0$

Step 2.3

Set ${(x - 3)}^{5}$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set ${(x - 3)}^{5}$ equal to $0$ .

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

Step 2.3.2

Solve ${(x - 3)}^{5} = 0$ for $x$ .

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Step 2.3.2.1

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

$x - 3 = 0$

Step 2.3.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.4

Set ${(x + 4)}^{5}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set ${(x + 4)}^{5}$ equal to $0$ .

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

Step 2.4.2

Solve ${(x + 4)}^{5} = 0$ for $x$ .

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Step 2.4.2.1

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.4.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.5

Set ${(x - 1 + \sqrt{7})}^{3}$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set ${(x - 1 + \sqrt{7})}^{3}$ equal to $0$ .

${(x - 1 + \sqrt{7})}^{3} = 0$

Step 2.5.2

Solve ${(x - 1 + \sqrt{7})}^{3} = 0$ for $x$ .

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Step 2.5.2.1

Set the $x - 1 + \sqrt{7}$ equal to $0$ .

$x - 1 + \sqrt{7} = 0$

Step 2.5.2.2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.5.2.2.1

Add $1$ to both sides of the equation.

$x + \sqrt{7} = 1$

Step 2.5.2.2.2

Subtract $\sqrt{7}$ from both sides of the equation.

$x = 1 - \sqrt{7}$

Step 2.6

The final solution is all the values that make ${(x^{2} + x - 12)}^{5} {(x - 1 + \sqrt{7})}^{3} = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = 3$ (Multiplicity of $5$ )

$x = - 4$ (Multiplicity of $5$ )

$x = 1 - \sqrt{7}$ (Multiplicity of $3$ )

$x = 3$ (Multiplicity of $5$ )

$x = - 4$ (Multiplicity of $5$ )

$x = 1 - \sqrt{7}$ (Multiplicity of $3$ )

Step 3