How do you simplify (2t^2+7t-4)/(-2t^2-5t+3)?

Jul 27, 2018

$\frac{t + 4}{- 1 \left(t + 3\right)}$

Explanation:

Factorize first as described below:

$2 {t}^{2} + 7 t - 4$

Factor by splitting the middle term

Step-1 : Multiply the coefficient of the first term by the constant $2 \times - 4 = - 8$

Step-2 : Find two factors of $- 8$ whose sum equals the coefficient of the middle term, which is $7$.

$- 1 + 8 = 7$

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, $- 1$ and $8$

$2 {t}^{2} - 1 t + 8 t - 4$

$t \left(2 t - 1\right) + 4 \left(2 t - 1\right)$

$\left(2 t - 1\right) \left(t + 4\right)$ -----> Factors!

Now lets do the same for $- 2 {t}^{2} - 5 t + 3$

Re-write the equation as $\textcolor{red}{- 1 \left(2 {t}^{2} + 5 t - 3\right)}$

Factor by splitting the middle term

Step-1 : Multiply the coefficient of the first term by the constant $2 \times - 3 = - 6$

Step-2 : Find two factors of $- 6$ whose sum equals the coefficient of the middle term, which is $5$ .

$- 1 + 6 = 5$

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step $2$ above, $- 1$ and $6$.

$2 {t}^{2} - 1 t + 6 t - 3$

$t \left(2 t - 1\right) + 3 \left(2 t - 1\right)$

$- 1 \left(2 t - 1\right) \left(t + 3\right)$ ----- Factors!

So now we get:

$\frac{2 {t}^{2} + 7 t - 4}{- 2 {t}^{2} - 5 t + 3}$ = ((2t-1)(t+4))/(-1(2t-2)(t+3)

((2t-1)(t+4))/(-1(2t-1)(t+3) = (cancel(2t-1)(t+4))/(-1cancel(2t-1)(t+3)

$\frac{t + 4}{- 1 \left(t + 3\right)}$