We assume that "P_2=2t^2+5t+4". "V" can be expressed in the following way: "V=c_1P_1+c_2P_2+c_3P_3" , "c_1,c_2,c_3\\in \\mathbb{C}" . Coefficients satisfy the following system:
"\\left\\{\\begin{matrix}\n c_1+2c_2+c_3=2 \\\\\n 2c_1+5c_2+3c_3=5 \\\\\nc_1+4c_2+6c_3=-5\n\\end{matrix}\\right."We multiply the first equation by "(-2)" and add to the second one. Then, we multiply the first equation by "(-1)" and add to the third one. As a result, we obtain we eliminate "c_1" from all equations and obtain the system for "c_2,c_3:"
"\\left\\{\\begin{matrix}\nc_2+c_3=1 \\\\\n2c_2+5c_3=-7\n\\end{matrix}\\right."
We multiply the first equation by "(-2)" and add to the second one. We obtain that "c_3=-3" . Then, from the first equation of the latter system we get: "c_2=4". From the first system we get "c_1=-3" . Thus, the solution is: "c_1=-3,c_2=4,c_3=-3" . The answer is: "V=-3P_1+4P_2-3P_3" .
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