# Polynomial problems

First problem:

Let $P \in \mathbb R[X], P(x) = x^3 + ax^2 + bx + c$ a polynomial with the roots $x_1, x_2, x_3, \:\:x_1 \neq x_2 \neq x_3$. For $Q \in \mathbb R[X]$ a first degree polynomial, the sum $$\frac{Q(x_1)}{P'(x_1)}+\frac{Q(x_2)}{P'(x_2)}+\frac{Q(x_3)}{P'(x_3)} = \: ?$$

I gave a form to $Q(x) = dx + e$ and replaced all in the sum. I don't know what to do next as the calculations are enormous.

Second problem:

I need to find the remainder of $X^{10} / (X+1)^2$ and $X^{10} / (X+1)^3$. I know how to find the remainder of something like $X^{10} / (X+1)$, which is $(-1)^{10}$, but not on those forms.

In your first problem, write $P(x) = (x-x_1)(x-x_2)(x-x_3)$ .

Then it has the following solution:

$$\frac{Q(x_1)}{P'(x_1)}+\frac{Q(x_2)}{P'(x_2)}+\frac{Q(x_3)}{P'(x_3)} = \\ \frac{Q(x_1)}{(x_1-x_2)(x_1-x_3)}+\frac{Q(x_2)}{(x_2-x_1) (x_2-x_3)}+\frac{Q(x_3)}{(x_3-x_1) (x_3-x_2)} =\\ \frac{(x_2-x_3)Q(x_1)}{(x_1-x_2)(x_1-x_3)(x_2-x_3)}-\frac{Q(x_2)(x_1-x_3)}{(x_1-x_2) (x_2-x_3)(x_1-x_3)}+\frac{Q(x_3)(x_1-x_2)}{(x_1-x_3) (x_2-x_3)(x_1-x_2)} =\\ \frac{1}{(x_1-x_3) (x_2-x_3)(x_1-x_2)}\Big[ x_1 (Q(x_3) - Q(x_2)) + x_2 (Q(x_1) - Q(x_3)) + x_3 (Q(x_2) - Q(x_1)) \Big] = \cdots$$

Let $Q(x) = dx + e$, then we have further

$$\cdots = \frac{d}{(x_1-x_3) (x_2-x_3)(x_1-x_2)}\Big[ x_1 (x_3 - x_2) + x_2 (x_1 - x_3) + x_3 (x_2 -x_1) \Big] = 0$$

• Why is $P'(x_1) = (x_1-x_2)(x_1-x_3)$ and so on for $x_2, x_3$ ? Commented Apr 17, 2017 at 7:35
• By the product rule of differentiation, you have $P'(x) = (x-x_1)(x-x_2) + (x-x_2)(x-x_3) + (x-x_1)(x-x_3)$ and hence $P'(x_1) = (x_1-x_2) (x_1-x_3)$. Likewise for the other terms. Commented Apr 17, 2017 at 10:45

Hint (for the second problem): the remainder of $x^{10} / (x+1)^2$ will be a linear polynomial $R(x)\,$, whose coefficients can be determined from the following conditions:

• $P(x) = x^{10} = (x+1)^2 \cdot Q(x) + R(x) \quad \implies \quad R(-1)=P(-1)$

• $P'(x) = 10\, x^{9} = 2(x+1) \cdot Q(x) + (x+1)^2 \cdot Q'(x) + R'(x) \quad \implies \quad R'(-1)=P'(-1)$