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

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Answer 1 · 2018-06-06T19:20:10

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

$[(cancel(t^2)+3t-8)-(cancel(t^2)-2t+6)] (t-4)$

$= (5t-14)(t-4)$

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

$= (5t-14)(t-4)$

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

How do you find the product [(t^2+3t-8)-(t^2-2t+6)](t-4)[(t2+3t−8)−(t2−2t+6)](t−4)[(t^2+3t-8)-(t^2-2t+6)](t-4)?