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

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

Step 1

Reorder terms.

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

Step 2

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

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

Step 3

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

To write $5$ as a fraction with a common denominator, multiply by $\frac{t^{2}}{t^{2}}$ .

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

Step 6

Combine the numerators over the common denominator.

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

Step 7

Factor by grouping.

Reorder terms.

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

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

Factor $- 1$ out of $- t$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $5 t + 4$ .

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

Step 8

Set the numerator equal to zero.

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

Step 9

Solve the equation for $t$ .

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

$5 t + 4 = 0$

$t - 1 = 0$

Set $5 t + 4$ equal to $0$ and solve for $t$ .

Set $5 t + 4$ equal to $0$ .

$5 t + 4 = 0$

Solve $5 t + 4 = 0$ for $t$ .

Subtract $4$ from both sides of the equation.

$5 t = - 4$

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

Divide each term in $5 t = - 4$ by $5$ .

$\frac{5 t}{5} = \frac{- 4}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 t}{5} = \frac{- 4}{5}$

Divide $t$ by $1$ .

$t = \frac{- 4}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{4}{5}$

Set $t - 1$ equal to $0$ and solve for $t$ .

Set $t - 1$ equal to $0$ .

$t - 1 = 0$

Add $1$ to both sides of the equation.

$t = 1$

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

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

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

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