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enter image description here

While I was doing some bases excercises, I checked the solution and found that they got that linear system above from $c_1 \cdot (x^2 -x+1) + c_2\cdot(2x+1)+c_3\cdot (2x-1)$

However, when I tried it myself, I got this as a linear system $$\begin{cases} c_1= a_2\\ -c_1+2c_2+2c_3=a_1\\ c_1+c_2-c_3=a_0 \end{cases}$$

What am I doing wrong here?

Edit : Forgot to add $a_0$ in the third equation. Fixed.

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    $\begingroup$ Your third line is not an equation. This is probably not right... $\endgroup$ – Yves Daoust Sep 25 '19 at 14:53
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    $\begingroup$ Maybe you forgot "$=a_0$" in the third line? $\endgroup$ – user Sep 25 '19 at 14:54
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Your are doing nothing wrong... The system you obtain is equivalent to the first one you mention, if you change the rhs, namely substituting $a_1$ by $a_1+a_2$ and $a_0$ by $a_0-a_2$. They probably just forgot to move $c_1$ to the rhs.

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From the expression given we obtain

$$c_1\cdot x^2+(-c_1+2c_2+2c_3)x+(c_1+c_2-c_3)$$

which implies

\begin{cases} c_1= a_2\\ -c_1+2c_2+2c_3=a_1\\ c_1+c_2-c_3=a_0 \end{cases}

which corresponds to your derivation.

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