# Factoring derivative of a polynomial using its roots

P(x) is a polynomial with roots x1,x2,x3...xn and P'(x) is its derivative. So can anyone explain how P'(x) can be factored as shown in the screenshot, using the roots of P(x)?!

• This follows from the usual product rule. Try writing it out for a cubic, you'll see the pattern.
– lulu
Jan 22, 2020 at 17:38

Let $$f_k(x) = x-x_k$$, note that $$f_k'(x) = 1$$.

Since $$P(x) = f_1(x) f_2(x)\cdots f_n(x)$$, the product differentiation rule gives $$P'(x) = f_1'(x) f_2(x)\cdots f_n(x) + f_1(x) f_2'(x)\cdots f_n(x) + \cdots + f_1(x) f_2(x)\cdots f_n'(x)$$, or $$P'(x) = f_2(x)\cdots f_n(x) + f_1(x) f_3(x)\cdots f_n(x) + \cdots + f_1(x) f_2(x)\cdots f_{n-1}(x)$$.

• Ohh I get it now, I just never used the product rule for more than 2 factors. Thanks!!! Jan 22, 2020 at 18:31

$$P(x)=\prod (x-x_i)$$ and so $$\ln P(x)=\sum \ln (x-x_i)$$

Differentiate w.r.t. $$x$$

$$\frac{P'(x)}{P(x)}=\sum \frac{1}{x-x_i}$$

So you want to prove $${P'(x)\over P(x) } = {1\over x-x_1}+{1\over x-x_2}+...+{1\over x-x_n}$$

but this is the same as $$(\ln P(x))' = (\ln(x-x_1)+\ln(x-x_2)+...+\ln(x-x_n))'$$

or $$(\ln P(x))' = (\ln(x-x_1)(x-x_2)...(x-x_n))'$$ or $$\ln P(x) = \ln(x-x_1)(x-x_2)...(x-x_n)+const$$

If we write $$a= e^{const}$$ i.e. $$const = \ln a$$ we get $$p(x) =a(x-x_1)(x-x_2)...(x-x_n)$$

Now read this from down to up and thus a conslusion.