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 Question from Christy, a student: Hi guys, ok so I'm having problems with solving this equation since it's been so long I've done precal. I have to find critical points so I have to find the derivative first. This is what I've done so far s(t)= (t-6)^4 * (t+1)^2 t'= 4(t-6)^3*(1)*(t+1)^2+ 2(t+1)(1)*(t-6)^4 so this is where I got stuck, which is the algebra part. How can I simplify this to get the answer to figure out the critical points.

Hi Christy,

I agree with your differentiation. If $s(t) = (t - 6)^4 (t + 1)^2$ then

$s'(t) = 4(t - 6)^3 (t + 1)^2 + 2(t + 1) (t - 6)^4 .$

This expression has two terms $4(t - 6)^3 (t + 1)^2 \mbox{ and } 2(t + 1) (t - 6)^4$ and these terms have common factors, $2, (t - 6)^3$ and $(t + 1).$ Hence the derivative $s'(t)$ can be written as

$s'(t) = 2 (t - 6)^3 (t + 1)[2(t + 1) + (t - 6)].$

Can you proceed from here?

Penny

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