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### Section 1-1 : Review : Functions

15. Determine all the roots of $$f\left( t \right) = {t^{\frac{5}{3}}} - 7{t^{\frac{4}{3}}} - 8t$$.

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Set the function equal to zero and factor the left side as much as possible.

${t^{\frac{5}{3}}} - 7{t^{\frac{4}{3}}} - 8t = t\left( {{t^{\frac{2}{3}}} - 7{t^{\frac{1}{3}}} - 8} \right) = t\left( {{t^{\frac{1}{3}}} - 8} \right)\left( {{t^{\frac{1}{3}}} + 1} \right) = 0$

Don’t so locked into quadratic equations that the minute you see an equation that is not quadratic you decide you can’t deal with it. While this function was not a quadratic it still factored, it just wasn’t as obvious that it did in this case. You could have clearly seen that if factored if it had been,

$t\left( {{t^2} - 7t - 8} \right)$

but notice that the only real difference is that the exponents are fractions now, but it still has the same basic form and so can be factored.

Okay, back to the problem. From the factored form we get,

\begin{align*} \begin{aligned}t & = 0 \\ t^{\frac{1}{3}} - 8 & = 0 \\ t^{\frac{1}{3}} + 1 & = 0 \end{aligned} & \begin{aligned} \\ \hspace{0.5in} \Rightarrow \hspace{0.25in} \\ \hspace{0.25in} \Rightarrow \hspace{0.25in} \end{aligned} & \begin{aligned} \\ t^{\frac{1}{3}} & = 8 \\ t^{\frac{1}{3}} & = - 1 \end{aligned} & \begin{aligned} \\ \hspace{0.5in} \Rightarrow \hspace{0.25in} \\ \hspace{0.25in} \Rightarrow \hspace{0.25in} \end{aligned} & \begin{aligned} \\ t & = {8^3} = 512 \\ t & = {\left( { - 1} \right)^3} = - 1 \end{aligned} \end{align*}

So, the function has three roots,

$t = - 1,\,\,\,\,\,t = 0,\,\,\,\,\,t = 512$