# How do you find the product [(t^2+3t-8)-(t^2-2t+6)](t-4)?

Jun 6, 2018

$\left[\left({t}^{2} + 3 t - 8\right) - \left({t}^{2} - 2 t + 6\right)\right] \left(t - 4\right) = 5 {t}^{2} - 34 t + 56$

#### Explanation:

$\left[\left(\cancel{{t}^{2}} + 3 t - 8\right) - \left(\cancel{{t}^{2}} - 2 t + 6\right)\right] \left(t - 4\right)$

$= \left(5 t - 14\right) \left(t - 4\right)$

We have the first term:
${t}^{2} + 3 t - 8 - {t}^{2} + 2 t + 6 = 5 t - 14$
so we get
$\left[\left(\cancel{{t}^{2}} + 3 t - 8\right) - \left(\cancel{{t}^{2}} - 2 t + 6\right)\right] \left(t - 4\right)$

$= \left(5 t - 14\right) \left(t - 4\right)$

$\left(5 t - 14\right) \left(t - 4\right) = 5 {t}^{2} - 14 t - 20 t + 56$
Combining like terms, we get:
$5 {t}^{2} - 34 t + 56$