Question #222357

Express V= 3t² + 7t + -4 as a linear combination of the polynomials

P1= t² + 2t + 3

P2= 2t² + 3t + 7

P3 = 3t² + 5t + 6


1
Expert's answer
2021-08-02T14:46:08-0400

Let us express V=3t2+7t4V= 3t^2+ 7t -4 as a linear combination of the polynomials

P1=t2+2t+3, P2=2t2+3t+7, P3=3t2+5t+6.P_1= t^2 + 2t + 3,\ P_2= 2t^2 + 3t + 7,\ P_3 = 3t^2 + 5t + 6.


Let V=aP1+bP2+cP3.V=aP_1+bP_2+cP_3.

Then

3t2+7t4=a(t2+2t+3)+b(2t2+3t+7)+c(3t2+5t+6).3t^2+ 7t -4= a(t^2 + 2t + 3)+b( 2t^2 + 3t + 7)+c( 3t^2 + 5t + 6).

It follows that

3t2+7t4=at2+2at+3a+2bt2+3bt+7b+3ct2+5ct+6c,3t^2+ 7t -4= at^2 + 2at + 3a+2bt^2 + 3bt + 7b+3ct^2 + 5ct + 6c,

and hence

3t2+7t4=(a+2b+3c)t2+(2a+3b+5c)t+(3a+7b+6c).3t^2+ 7t -4= (a+2b+3c)t^2 + (2a+3b+5c)t + (3a+ 7b + 6c).


Therefore, we have the following system:


{a+2b+3c=32a+3b+5c=73a+7b+6c=4\begin{cases} a+2b+3c=3\\ 2a+3b+5c=7\\ 3a+ 7b + 6c=-4 \end{cases}


Let us multiply the first equation by 2-2 and add to the second equation, also multiply the first equation by 3-3 and add to the third equation. Thgen we have the system:


{a+2b+3c=3bc=1b3c=13\begin{cases} a+2b+3c=3\\ -b-c=1\\ b -3c=-13 \end{cases}


Let us add the last to equation. Then we get the system:


{a=32b3cbc=14c=12\begin{cases} a=3-2b-3c\\ -b-c=1\\ -4c=-12 \end{cases}


The last system is equivalent to the system:


{a=2b=4c=3\begin{cases} a=2\\ b=-4\\ c=3 \end{cases}


We conclude that V=2P14P2+3P3.V=2P_1-4P_2+3P_3.


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