Solve by Factoring x^-2+2x^-1-15=0

$x^{- 2} + 2 x^{- 1} - 15 = 0$

Step 1

Factor $x^{- 2} + 2 x^{- 1} - 15$ using the AC method.

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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(x^{- 1} - 3) (x^{- 1} + 5) = 0$

Step 2

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

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

Step 3

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

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

Step 4

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

$(\frac{1}{x} - 3 \cdot \frac{x}{x}) (\frac{1}{x} + 5) = 0$

Step 5

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

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

Step 6

Combine the numerators over the common denominator.

$\frac{1 - 3 x}{x} \cdot (\frac{1}{x} + 5) = 0$

Step 7

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

$\frac{1 - 3 x}{x} \cdot (\frac{1}{x} + \frac{5 x}{x}) = 0$

Step 8

Combine the numerators over the common denominator.

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

Step 9

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

$\frac{1 - 3 x}{x} = 0$

$\frac{1 + 5 x}{x} = 0$

Step 10

Set $\frac{1 - 3 x}{x}$ equal to $0$ and solve for $x$ .

Set $\frac{1 - 3 x}{x}$ equal to $0$ .

$\frac{1 - 3 x}{x} = 0$

Solve $\frac{1 - 3 x}{x} = 0$ for $x$ .

Set the numerator equal to zero.

$1 - 3 x = 0$

Solve the equation for $x$ .

Subtract $1$ from both sides of the equation.

$- 3 x = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

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

Step 11

Set $\frac{1 + 5 x}{x}$ equal to $0$ and solve for $x$ .

Set $\frac{1 + 5 x}{x}$ equal to $0$ .

$\frac{1 + 5 x}{x} = 0$

Solve $\frac{1 + 5 x}{x} = 0$ for $x$ .

Set the numerator equal to zero.

$1 + 5 x = 0$

Solve the equation for $x$ .

Subtract $1$ from both sides of the equation.

$5 x = - 1$

Divide each term in $5 x = - 1$ by $5$ and simplify.

Divide each term in $5 x = - 1$ by $5$ .

$\frac{5 x}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 1}{5}$

Divide $x$ by $1$ .

$x = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{5}$

Step 12

The final solution is all the values that make $\frac{1 - 3 x}{x} \cdot \frac{1 + 5 x}{x} = 0$ true.

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

Solve by Factoring x^-2+2x^-1-15=0

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