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Improved part to this question

Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define $$ \langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1). $$

(i) Prove that $\langle\cdot,\cdot\rangle$ defines an inner product.

Here is what I had:

Consider $P_t$, the space of polynomial of degree polynomial at most $t$ ( with coefficients from R)Let $x_0<x_1<x_2<…<x_3$ be any $t+1$ real numbers. For $p$ and q belong to $P_t$ ,we define $<p,q>$ to be $p(1)q(1)+p(1)'q(1)'+p(1)''q(1)''+p(1)'''q(1)'''$ is an inner product on $P_t$. Checking that $<p,P> = 0 $->$ p = 0$ then $1$ is the root of $p$, and as a polynomial of degree at most $t$ has at most $t$ roots unless it's the zero polynomial, p must be the zero polynomial. Different choice of $1$ gives rise to different inner products.

Could we just do this $f(a)=0$ and $f'(a)=0$ then $f(x)$ is divisible by $(x-a)^2$ ?

Can someone please help me with this proof for part (i). It is frustrating me.

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$$ \langle p,p\rangle = p(1)^2 + p'(1)^2 + p''(1)^2 + p'''(1)^2 \geq 0 $$ And if $\langle p,p\rangle = 0$, then note that $$ p(1) = p'(1) = p''(1) = p'''(1) = 0 $$ Now write $$ p(t) = at^3 + bt^2 + ct + d $$ and see that $a=b=c=d=0$ and conclude that $$ p = 0 $$

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