Question

Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(x), T(x2), and T(ax2+bx+c), where a, b, and c are arbitrary real numbers.

Answer 1

It looks like we're told that

$T(-2x^2)=3x^2+4x$

$T\left(\dfrac12x+3\right)=-3x^2+3x-4$

$T(2x^2-1)=-3x+4$

We use the fact that $T$ is linear to find $T(1),T(x),T(x^2)$. First, we notice that

$T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)$

We also have

$T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)$

$T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)$

So once we find $T(1)$, we can determine $T(x)$ and $T(x^2)$. We have

$T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4$

and using this we find

$T(x)=12x^2+12x+16$

$T(x^2)=-\dfrac32x^2-2x$

Then

$T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)$

$T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c$