Solve by Factoring 4-1/x-5/(x^2)=0

$4 - \frac{1}{x} - \frac{5}{x^{2}} = 0$

Step 1

Reorder terms.

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

Step 2

To write $- \frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 3

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

Multiply $\frac{1}{x}$ by $\frac{x}{x}$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

To write $4$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Factor by grouping.

Reorder terms.

$\frac{4 x^{2} - x - 5}{x^{2}} = 0$

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 $a \cdot c = 4 \cdot - 5 = - 20$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$\frac{4 x^{2} - (x) - 5}{x^{2}} = 0$

Rewrite $- 1$ as $4$ plus $- 5$

$\frac{4 x^{2} + (4 - 5) x - 5}{x^{2}} = 0$

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(4 x^{2} + 4 x) - 5 x - 5}{x^{2}} = 0$

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

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

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

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

Step 8

Set the numerator equal to zero.

$(x + 1) (4 x - 5) = 0$

Step 9

Solve the equation for $x$ .

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

$x + 1 = 0$

$4 x - 5 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $4 x - 5$ equal to $0$ and solve for $x$ .

Set $4 x - 5$ equal to $0$ .

$4 x - 5 = 0$

Solve $4 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$4 x = 5$

Divide each term in $4 x = 5$ by $4$ and simplify.

Divide each term in $4 x = 5$ by $4$ .

$\frac{4 x}{4} = \frac{5}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{5}{4}$

Divide $x$ by $1$ .

$x = \frac{5}{4}$

The final solution is all the values that make $(x + 1) (4 x - 5) = 0$ true.

$x = - 1, \frac{5}{4}$

Solve by Factoring 4-1/x-5/(x^2)=0

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