Simplify f(t)=(2t(t+1))/(t-1)-(3+t)/(t-1)

$f (t) = \frac{2 t (t + 1)}{t - 1} - \frac{3 + t}{t - 1}$

Combine the numerators over the common denominator.

$f (t) = \frac{2 t (t + 1) - (3 + t)}{t - 1}$

Simplify each term.

Apply the distributive property.

$f (t) = \frac{2 t \cdot t + 2 t \cdot 1 - (3 + t)}{t - 1}$

Multiply $t$ by $t$ by adding the exponents.

Move $t$ .

$f (t) = \frac{2 (t \cdot t) + 2 t \cdot 1 - (3 + t)}{t - 1}$

Multiply $t$ by $t$ .

$f (t) = \frac{2 t^{2} + 2 t \cdot 1 - (3 + t)}{t - 1}$

Multiply $2$ by $1$ .

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

Apply the distributive property.

$f (t) = \frac{2 t^{2} + 2 t - 1 \cdot 3 - t}{t - 1}$

Multiply $- 1$ by $3$ .

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

Subtract $t$ from $2 t$ .

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

Factor by grouping.

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$f (t) = \frac{2 t (t - 1) + 3 (t - 1)}{t - 1}$

Factor the polynomial by factoring out the greatest common factor, $t - 1$ .

$f (t) = \frac{(t - 1) (2 t + 3)}{t - 1}$

Cancel the common factor of $t - 1$ .

Cancel the common factor.

$f (t) = \frac{(t - 1) (2 t + 3)}{t - 1}$

Divide $2 t + 3$ by $1$ .

$f (t) = 2 t + 3$

Simplify f(t)=(2t(t+1))/(t-1)-(3+t)/(t-1)

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