Calculus 1 polynomials and derivatives

Question

Need your help with a Calculus 1 exercise.

Let $P,Q$ be polyomials with $deg(P)

Answer 1

Hint

See that

$$P(x)= \frac{A_1Q(x)}{x-a_1}+h(x)$$

But

$$P(a_1)=A_1(a_1-a_2)(a_1-a_3)...(a_1-a_n)+h(a_1)$$

But is easy to see that $h(a_1)=0$ and we also have $(a_1-a_2)(a_1-a_3)...(a_1-a_n)=Q'(a_1)$.

Can you finish?

Answer 2

Let us suppose that $P,Q$ are polynomials with real coefficients and that $Q=(X-a_1)(X-a_2)\cdots(X-a_n)$ where the $a_i$ are pairwise distinct real numbers.

The theorem which gives the existence (and uniqueness) of partial fractions decomposition asserts that there exist real coefficients $A_1,\cdots,A_n$ such that :

$$\frac{P}{Q}=\frac{A_1}{X-a_1}+\cdots+\frac{A_n}{X-a_n}$$

Now, multiplying both sides by $Q$ and evaluating at $a_j$ for some $j\in\{1,\cdots,n\}$ leads to :

$$P(a_j)=A_j\prod_{i\neq j}(a_j-a_i)$$

But with know that :

$$Q'=\sum_{k=1}^n\prod_{i\neq k}(X-a_i)$$ hence :

$$Q'(a_j)=\prod_{i\neq j}(a_j-a_i)$$

Finally :

$$\boxed{\frac{P}{Q}=\sum_{i=1}^n\frac{\frac{P(a_i)}{Q'(a_i)}}{X-a_i}}$$