Pre-Algebra Examples

$\frac{x^{2} - 2 x + 1}{x^{2} + 2 x - 24} > 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - 2 x + 1 = 0$

$x^{2} + 2 x - 24 = 0$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $1$ as $1^{2}$ .

$x^{2} - 2 x + 1^{2} = 0$

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 2.3

Rewrite the polynomial.

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

Step 2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 3

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Factor $x^{2} + 2 x - 24$ using the AC method.

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Step 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 5.2

Write the factored form using these integers.

$(x - 4) (x + 6) = 0$

Step 6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 6 = 0$

Step 7

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 7.2

Add $4$ to both sides of the equation.

$x = 4$

Step 8

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 8.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 9

The final solution is all the values that make $(x - 4) (x + 6) = 0$ true.

$x = 4, - 6$

Step 10

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1$

$x = 4, - 6$

Step 11

Consolidate the solutions.

$x = 1, 4, - 6$

Step 12

Find the domain of $\frac{x^{2} - 2 x + 1}{x^{2} + 2 x - 24}$ .

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Step 12.1

Set the denominator in $\frac{x^{2} - 2 x + 1}{x^{2} + 2 x - 24}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 24 = 0$

Step 12.2

Solve for $x$ .

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Step 12.2.1

Factor $x^{2} + 2 x - 24$ using the AC method.

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Step 12.2.1.1

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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 12.2.1.2

Write the factored form using these integers.

$(x - 4) (x + 6) = 0$

Step 12.2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 6 = 0$

Step 12.2.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 12.2.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 12.2.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 12.2.4

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 12.2.4.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 12.2.4.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 12.2.5

The final solution is all the values that make $(x - 4) (x + 6) = 0$ true.

$x = 4, - 6$

Step 12.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 6) \cup (- 6, 4) \cup (4, \infty)$

Step 13

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 1$

$1 < x < 4$

$x > 4$

Step 14

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 14.1

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

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Step 14.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 14.1.2

Replace $x$ with $- 8$ in the original inequality.

$\frac{{(- 8)}^{2} - 2 \cdot - 8 + 1}{{(- 8)}^{2} + 2 (- 8) - 24} > 0$

Step 14.1.3

The left side $3.375$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 14.2

Test a value on the interval $- 6 < x < 1$ to see if it makes the inequality true.

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Step 14.2.1

Choose a value on the interval $- 6 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 14.2.2

Replace $x$ with $0$ in the original inequality.

$\frac{{(0)}^{2} - 2 \cdot 0 + 1}{{(0)}^{2} + 2 (0) - 24} > 0$

Step 14.2.3

The left side $- 0.041\overline{6}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 14.3

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

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Step 14.3.1

Choose a value on the interval $1 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 14.3.2

Replace $x$ with $2$ in the original inequality.

$\frac{{(2)}^{2} - 2 \cdot 2 + 1}{{(2)}^{2} + 2 (2) - 24} > 0$

Step 14.3.3

The left side $- 0.0625$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 14.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 14.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 14.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{{(6)}^{2} - 2 \cdot 6 + 1}{{(6)}^{2} + 2 (6) - 24} > 0$

Step 14.4.3

The left side $1.041\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 14.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 6$ True

$- 6 < x < 1$ False

$1 < x < 4$ False

$x > 4$ True

$x < - 6$ True

$- 6 < x < 1$ False

$1 < x < 4$ False

$x > 4$ True

Step 15

The solution consists of all of the true intervals.

$x < - 6$ or $x > 4$

Step 16

The result can be shown in multiple forms.

Inequality Form:

$x < - 6 or x > 4$

Interval Notation:

$(- \infty, - 6) \cup (4, \infty)$

Step 17