How do you find $w (t - 1)$ $w(t-1)$ given $w (t) = t^{2} - 4$ $w(t)=t^2-4$ ?

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Answer 1 · 2017-01-15T11:02:17

The answer is $= (t + 1) (t - 3)$ $=(t+1)(t-3)$

Replace $t$ $t$ by $(t - 1)$ $(t-1)$

$w (t) = t^{2} - 4$ $w(t)=t^2-4$

$w (t - 1) = {(t - 1)}^{2} - 4 = t^{2} - 2 t + 1 - 4$ $w(t-1)=(t-1)^2-4=t^2-2t+1-4$

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

How do you find w(t−1)w(t-1) given w(t)=t2−4w(t)=t^2-4?