I need help on the following question. I have no idea how to attempt it.
Find the dimension of the vector space: $$ {p(x) = ax^3 + bx^2 + cx + d \in P_3:p'(1) = 0}. $$
$\endgroup$2 Answers
$\begingroup$\begin{align} p'(x) &= 3ax^2 + 2bx + c \\ p'(1) &= 0 \iff 3a + 2b + c = 0 \iff c = -(3a + 2b) \end{align}
This means that you lose a degree of freedom, and subspace contains polynomials of the form $$p(x) = ax^3 +bx^2 - (3a+2b)x + d.$$
Can you figure out the dimension from this?
$\endgroup$ 2 $\begingroup$The dimension of Pn(F) is n+1 so for your P3, the dim=4.
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