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Let T:P3→P3 be the linear transformation such that $T(−2x^2)= −2x^2 − 2x$, $T(0.5x + 2)= 3x^2 + 4x−2$, and $T(2x^2 − 1)= 2x + 1$. Find $T(1), T(x), T(x^2)$, and $T(ax^2 + bx + c)$, where a, b, and c are arbitrary real numbers.

I understand how to find $T(x^2)$ where you just divide the given $T(-2x^2)$ by $-2$ to get $T(x^2) = x^2 + x$.

I'm not sure sure how to proceed in calculating the other transformation functions. Please list as many of the steps as possible in solving for the three other functions.

Thank you!

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We can write $1=-(-2x^2)-(2x^2-1), x=2(0.5x+2)-4$, so $$T(1)=T(-(-2x^2)-(2x^2-1))=-T(-2x^2)-T(2x^2-1)=2x^2-1$$ $$T(x)=T(2(0.5x+2)-4)=2T(0.5x+2)-4T(1)=-2x^2+8x$$ Then we get $$T(ax^2+bx+c)=a(x^2+x)+b(-2x^2+8x)+c(2x^2-1)$$ $$=(a-2b+2c)x^2+(a+8b)x-c$$

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  • $\begingroup$ Oh! so you're making linear combinations of the portions inside the the parenthesis to equate to the unknown. That's such a more helpful way of looking at these kind of problems. Thank you. $\endgroup$ – Laughing Horse Nov 15 '15 at 21:01

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