91影视

If f is redundant if it is a linear combination of$f_1,f_2,...,f_{i-1}$then the elements$f_1,f_2,...,f_n$are linearly independent.

In this case the equation becomes

$c_1{f}_1+c_2{f}_2+...+c_n{f}_n=0$

Only has trivial solution

.$c_1=c_2=....=c_n=0$

Consider, the given polynomial as follows$f\left(t\right)=t+1\text{鈥�}\&$, $g\left(t\right)=(t+2)(t+k)$where k is an arbitrary constant.

$\begin{array}{c}f\left(t\right)+g\left(t\right)=(t+1)+(t+2)(t+k)\\ \text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}=(t+1)+(t^2+tk+2t+2k)\text{鈥夆赌夆赌�}\\ \text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌�}=(t+1)+t^2+(k+2)t+2k\end{array}$

Further calculation are as

$\begin{array}{l}=t^2+(k+2+1)t+(2k+1)\\ =t^2+(k+3)t+(2k+1)\end{array}$

Here, the values of polynomials are as follows

$\begin{array}{l}f\left(t\right)=t+1\\ tf\left(t\right)=t^2+t\\ g\left(t\right)=t^2+(k+2)t+2k\end{array}$

Therefore, the values of the arbitrary constant k are$鈭�\frac{1}{2},鈭�3$

Since, the polynomials $tf\left(t\right)=t^2+t,g\left(t\right)=t^2+(k+2)t+2k$are the basis of$P_2$.

Hence the solution.