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Let $T:P_3 \to P_3$ be the linear transformation such that $T(2 x^2)= -2 x^2 - 4 x$, $T(-0.5 x - 5)= 2 x^2 + 4 x + 3$, and $T(2 x^2 - 1) = 4 x - 4.$ Find $T(1)$, $T(x)$, $T(x^2)$, and $T(a x^2 + b x + c)$, where $a$, $b$, and $c$ are arbitrary real numbers.

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What have you tried? What does your notation mean? –  Tobias Kildetoft Feb 20 '13 at 17:32
    
Can you express $1$ and $x^2$ as linear combinations of $2x^2$ and $2x^2-1$ and build it from there? –  Jyrki Lahtonen Feb 20 '13 at 17:36
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hint:T is linear transformation $T(a+bx+cx^2)=aT(1)+bT(x) +cT(x^2) $ $$T(2 x^2)= -2 x^2 - 4 x \to T( x^2)= - x^2 - 2 x$$$$T(-0.5 x - 5)= 2 x^2 + 4 x + 3\to-0.5T( x) - 5T(1)= 2 x^2 + 4 x + 3\to 16x^2+72x-46$$$$T(2 x^2 - 1) = 4 x - 4.\to T( x^2) - \frac12T(1) = 2 x - 2\to T(1)=-2 x^2 - 8x +4.$$

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