Solve by Factoring 4-1/t-3/(t^2)=0

$4 - \frac{1}{t} - \frac{3}{t^{2}} = 0$

Step 1

Reorder terms.

$- \frac{1}{t} - \frac{3}{t^{2}} + 4 = 0$

Step 2

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

$- \frac{1}{t} \cdot \frac{t}{t} - \frac{3}{t^{2}} + 4 = 0$

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 $\frac{t}{t}$ .

$- \frac{t}{t \cdot t} - \frac{3}{t^{2}} + 4 = 0$

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

$- \frac{t}{t \cdot t} - \frac{3}{t^{2}} + 4 = 0$

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

$- \frac{t}{t \cdot t} - \frac{3}{t^{2}} + 4 = 0$

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

$- \frac{t}{t^{1 + 1}} - \frac{3}{t^{2}} + 4 = 0$

Add $1$ and $1$ .

$- \frac{t}{t^{2}} - \frac{3}{t^{2}} + 4 = 0$

Step 4

Combine the numerators over the common denominator.

$\frac{- t - 3}{t^{2}} + 4 = 0$

Step 5

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

$\frac{- t - 3}{t^{2}} + \frac{4 t^{2}}{t^{2}} = 0$

Step 6

Combine the numerators over the common denominator.

$\frac{- t - 3 + 4 t^{2}}{t^{2}} = 0$

Step 7

Factor by grouping.

Reorder terms.

$\frac{4 t^{2} - t - 3}{t^{2}} = 0$

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 $a \cdot c = 4 \cdot - 3 = - 12$ and whose sum is $b = - 1$ .

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

$\frac{4 t^{2} - (t) - 3}{t^{2}} = 0$

Rewrite $- 1$ as $3$ plus $- 4$

$\frac{4 t^{2} + (3 - 4) t - 3}{t^{2}} = 0$

Apply the distributive property.

$\frac{4 t^{2} + 3 t - 4 t - 3}{t^{2}} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(4 t^{2} + 3 t) - 4 t - 3}{t^{2}} = 0$

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

$\frac{t (4 t + 3) - (4 t + 3)}{t^{2}} = 0$

Factor the polynomial by factoring out the greatest common factor, $4 t + 3$ .

$\frac{(4 t + 3) (t - 1)}{t^{2}} = 0$

Step 8

Set the numerator equal to zero.

$(4 t + 3) (t - 1) = 0$

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 $0$ .

$4 t + 3 = 0$

$t - 1 = 0$

Set $4 t + 3$ equal to $0$ and solve for $t$ .

Set $4 t + 3$ equal to $0$ .

$4 t + 3 = 0$

Solve $4 t + 3 = 0$ for $t$ .

Subtract $3$ from both sides of the equation.

$4 t = - 3$

Divide each term in $4 t = - 3$ by $4$ and simplify.

Divide each term in $4 t = - 3$ by $4$ .

$\frac{4 t}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 t}{4} = \frac{- 3}{4}$

Divide $t$ by $1$ .

$t = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{3}{4}$

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 $(4 t + 3) (t - 1) = 0$ true.

$t = - \frac{3}{4}, 1$

Solve by Factoring 4-1/t-3/(t^2)=0

Related Questions

Support LCMGCF.com by disabling your adblocker.