Answer
$S$ is a basis for $P_3$. .
Work Step by Step
The set $S=\left\{t^{3}-2 t^{2}+1, t^{2}-4, t^{3}+2 t, 5 t\right\}$ is a basis for $P_3$. Indeed, consider the following linear combination
$$a(t^{3}-2 t^{2}+1)+b( t^{2}-4)+c(t^{3}+2 t)+d ( 5 t)=0.$$
Comparing the coefficients, we have
\begin{align*}
a-4b&=0\\
2c+5d&=0\\
-2a+b&=0\\
a+c&=0.
\end{align*}
The augmented matrix of the above system is given by
$$\left[\begin{array}{cccc}
{1}&{-4}&{0}&{0}\\
{0}&{0}&{2}&{5}\\
{-2}&{1}&{0}&{0}\\
{1}&{0}&{1}&{0}\\
\end{array}\right]$$
and its determinant is -35. Therefore, the above system has unique solution, that is, the trivial solution. Then $S$ is linearly independent set of vectors and since $P_3$ has dimension $4$, then by Theorem 4.12, $S$ is a basis for $P_3$.
