Let us express the following polynomial as a linear combination of the polynomials.
P = a P 1 + b P 2 + c P 3 P=aP_1+bP_2+cP_3 P = a P 1 + b P 2 + c P 3
It follows that
t 2 + 4 t − 3 = a ( t 2 + 2 t + 5 ) + b ( 2 t 2 − 3 t ) + c ( t + 3 ) t^2+ 4t-3= a(t^2+2t+5)+b(2t^2-3t)+c(t+3) t 2 + 4 t − 3 = a ( t 2 + 2 t + 5 ) + b ( 2 t 2 − 3 t ) + c ( t + 3 ) ,
and hence
t 2 + 4 t − 3 = ( a + 2 b ) t 2 + ( 2 a − 3 b + c ) t + ( 5 a + 3 c ) t^2+ 4t-3= (a+2b)t^2+(2a-3b+c)t+(5a+3c) t 2 + 4 t − 3 = ( a + 2 b ) t 2 + ( 2 a − 3 b + c ) t + ( 5 a + 3 c )
We get the following system:
{ a + 2 b = 1 2 a − 3 b + c = 4 5 a + 3 c = − 3 \begin{cases} a+2b=1\\2a-3b+c=4\\5a+3c=-3\end{cases} ⎩ ⎨ ⎧ a + 2 b = 1 2 a − 3 b + c = 4 5 a + 3 c = − 3
{ a = 1 − 2 b 2 ( 1 − 2 b ) − 3 b + c = 4 5 ( 1 − 2 b ) + 3 c = − 3 \begin{cases} a=1-2b\\2(1-2b)-3b+c=4\\5(1-2b)+3c=-3\end{cases} ⎩ ⎨ ⎧ a = 1 − 2 b 2 ( 1 − 2 b ) − 3 b + c = 4 5 ( 1 − 2 b ) + 3 c = − 3
{ a = 1 − 2 b − 7 b + c = 2 − 10 b + 3 c = − 8 \begin{cases} a=1-2b\\-7b+c=2\\-10b+3c=-8\end{cases} ⎩ ⎨ ⎧ a = 1 − 2 b − 7 b + c = 2 − 10 b + 3 c = − 8
{ a = 1 − 2 b c = 2 + 7 b − 10 b + 3 ( 2 + 7 b ) = − 8 \begin{cases} a=1-2b\\c=2+7b\\-10b+3(2+7b)=-8\end{cases} ⎩ ⎨ ⎧ a = 1 − 2 b c = 2 + 7 b − 10 b + 3 ( 2 + 7 b ) = − 8
{ a = 1 − 2 b c = 2 + 7 b 11 b = − 14 \begin{cases} a=1-2b\\c=2+7b\\11b=-14\end{cases} ⎩ ⎨ ⎧ a = 1 − 2 b c = 2 + 7 b 11 b = − 14
{ a = 39 11 c = − 76 11 b = − 14 11 \begin{cases} a=\frac{39}{11}\\c=-\frac{76}{11}\\b=-\frac{14}{11}\end{cases} ⎩ ⎨ ⎧ a = 11 39 c = − 11 76 b = − 11 14
We conclude that
P = 39 11 P 1 − 14 11 P 2 − 76 11 P 3 P=\frac{39}{11}P_1-\frac{14}{11}P_2-\frac{76}{11}P_3 P = 11 39 P 1 − 11 14 P 2 − 11 76 P 3
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