Question #211641

Express the following polynomial as a linear combination of the polynomials.


P=t2+ 4t-3

P1=t2+2t+5

P2=2t2-3t

P3=t+3


1
Expert's answer
2021-06-29T12:34:08-0400

Let us express the following polynomial as a linear combination of the polynomials.


P=aP1+bP2+cP3P=aP_1+bP_2+cP_3


It follows that


t2+4t3=a(t2+2t+5)+b(2t23t)+c(t+3)t^2+ 4t-3= a(t^2+2t+5)+b(2t^2-3t)+c(t+3),


and hence


t2+4t3=(a+2b)t2+(2a3b+c)t+(5a+3c)t^2+ 4t-3= (a+2b)t^2+(2a-3b+c)t+(5a+3c)


We get the following system:


{a+2b=12a3b+c=45a+3c=3\begin{cases} a+2b=1\\2a-3b+c=4\\5a+3c=-3\end{cases}


{a=12b2(12b)3b+c=45(12b)+3c=3\begin{cases} a=1-2b\\2(1-2b)-3b+c=4\\5(1-2b)+3c=-3\end{cases}


{a=12b7b+c=210b+3c=8\begin{cases} a=1-2b\\-7b+c=2\\-10b+3c=-8\end{cases}


{a=12bc=2+7b10b+3(2+7b)=8\begin{cases} a=1-2b\\c=2+7b\\-10b+3(2+7b)=-8\end{cases}


{a=12bc=2+7b11b=14\begin{cases} a=1-2b\\c=2+7b\\11b=-14\end{cases}


{a=3911c=7611b=1411\begin{cases} a=\frac{39}{11}\\c=-\frac{76}{11}\\b=-\frac{14}{11}\end{cases}


We conclude that


P=3911P11411P27611P3P=\frac{39}{11}P_1-\frac{14}{11}P_2-\frac{76}{11}P_3



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