Algebra Examples

$2 x^{- 2} - 5 x^{- 1} - 3 = 0$

Step 1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 (\frac{1}{x^{2}}) - 5 x^{- 1} - 3 = 0$

Step 2

Combine $2$ and $\frac{1}{x^{2}}$ .

$\frac{2}{x^{2}} - 5 x^{- 1} - 3 = 0$

Step 3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{2}{x^{2}} - 5 \frac{1}{x} - 3 = 0$

Step 4

Combine $- 5$ and $\frac{1}{x}$ .

$\frac{2}{x^{2}} + \frac{- 5}{x} - 3 = 0$

Step 5

Move the negative in front of the fraction.

$\frac{2}{x^{2}} - \frac{5}{x} - 3 = 0$

Step 6

Reorder terms.

$- \frac{5}{x} + \frac{2}{x^{2}} - 3 = 0$

Step 7

To write $- \frac{5}{x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$- \frac{5}{x} \cdot \frac{x}{x} + \frac{2}{x^{2}} - 3 = 0$

Step 8

Write each expression with a common denominator of $x^{2}$ , by multiplying each by an appropriate factor of $1$ .

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Step 8.1

Multiply $\frac{5}{x}$ by $\frac{x}{x}$ .

$- \frac{5 x}{x \cdot x} + \frac{2}{x^{2}} - 3 = 0$

Step 8.2

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

$- \frac{5 x}{x \cdot x} + \frac{2}{x^{2}} - 3 = 0$

Step 8.3

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

$- \frac{5 x}{x \cdot x} + \frac{2}{x^{2}} - 3 = 0$

Step 8.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{5 x}{x^{1 + 1}} + \frac{2}{x^{2}} - 3 = 0$

Step 8.5

Add $1$ and $1$ .

$- \frac{5 x}{x^{2}} + \frac{2}{x^{2}} - 3 = 0$

Step 9

Combine the numerators over the common denominator.

$\frac{- 5 x + 2}{x^{2}} - 3 = 0$

Step 10

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

$\frac{- 5 x + 2}{x^{2}} - 3 \cdot \frac{x^{2}}{x^{2}} = 0$

Step 11

Combine $- 3$ and $\frac{x^{2}}{x^{2}}$ .

$\frac{- 5 x + 2}{x^{2}} + \frac{- 3 x^{2}}{x^{2}} = 0$

Step 12

Combine the numerators over the common denominator.

$\frac{- 5 x + 2 - 3 x^{2}}{x^{2}} = 0$

Step 13

Factor by grouping.

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Step 13.1

Reorder terms.

$\frac{- 3 x^{2} - 5 x + 2}{x^{2}} = 0$

Step 13.2

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 $a \cdot c = - 3 \cdot 2 = - 6$ and whose sum is $b = - 5$ .

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Step 13.2.1

Factor $- 5$ out of $- 5 x$ .

$\frac{- 3 x^{2} - 5 x + 2}{x^{2}} = 0$

Step 13.2.2

Rewrite $- 5$ as $1$ plus $- 6$

$\frac{- 3 x^{2} + (1 - 6) x + 2}{x^{2}} = 0$

Step 13.2.3

Apply the distributive property.

$\frac{- 3 x^{2} + 1 x - 6 x + 2}{x^{2}} = 0$

Step 13.2.4

Multiply $x$ by $1$ .

$\frac{- 3 x^{2} + x - 6 x + 2}{x^{2}} = 0$

Step 13.3

Factor out the greatest common factor from each group.

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Step 13.3.1

Group the first two terms and the last two terms.

$\frac{(- 3 x^{2} + x) - 6 x + 2}{x^{2}} = 0$

Step 13.3.2

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

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

Step 13.4

Factor the polynomial by factoring out the greatest common factor, $- 3 x + 1$ .

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

Step 14

Set the numerator equal to zero.

$(- 3 x + 1) (x + 2) = 0$

Step 15

Solve the equation for $x$ .

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Step 15.1

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

$- 3 x + 1 = 0$

$x + 2 = 0$

Step 15.2

Set $- 3 x + 1$ equal to $0$ and solve for $x$ .

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Step 15.2.1

Set $- 3 x + 1$ equal to $0$ .

$- 3 x + 1 = 0$

Step 15.2.2

Solve $- 3 x + 1 = 0$ for $x$ .

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Step 15.2.2.1

Subtract $1$ from both sides of the equation.

$- 3 x = - 1$

Step 15.2.2.2

Divide each term in $- 3 x = - 1$ by $- 3$ and simplify.

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Step 15.2.2.2.1

Divide each term in $- 3 x = - 1$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 15.2.2.2.2

Simplify the left side.

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Step 15.2.2.2.2.1

Cancel the common factor of $- 3$ .

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Step 15.2.2.2.2.1.1

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 1}{- 3}$

Step 15.2.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 3}$

Step 15.2.2.2.3

Simplify the right side.

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Step 15.2.2.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{3}$

Step 15.3

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 15.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 15.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 15.4

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

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