# taylor expansion – Figuring out pairwise product of roots of a seventh polynomial given two remainder of the polynomial Let $$f(x)$$ be a polynomial of degree 7, which leaves a remainder of $$-1$$ and $$1$$ upon division by $$(x-1)^4$$ and $$(x+1)^4$$ respectively. What is the sum of all the pairwise product of roots / coefficient of x^5 divided by coefficient of $$x^7$$

My attempt:

I tried a bit of “different” approach to this question by using Taylor’s theorem, what I did was, I expanded $$f(x)$$ around $$x=1$$ and $$x=-1$$ and divided the respective polynomials by the factors to find remainder. I have shown the mathematics below:

$$frac{f(x)}{(x-a)^4} = frac{ sum_{k=0}^{k=7} frac{ f^{k} (1) }{k!} (x-a)^k }{ (x-a)^4} = sum_{k=4}^{k=7} frac{ f^{k} (1) (x-a)^{k-4} }{k!} + sum_{k=0}^{k=3} frac{ f^{k} (1) (x-a)^{k-4} }{k!}$$

The remainder is precisely the first sum:

$$-1 = sum_{k=4}^{k=7} frac{ f^{k} (1) (x-1)^{k-4} } {k!}$$

Note: I brought in condition that remainder is minus one

Similarly I get another result for Taylor expanding around the zero of the other quadratic factor:

$$1 = sum_{k=4}^{k=7} frac{ f^{k} (-1) (x+1)^{k-4} } {k!}$$

Expanding both, I get:

$$-1 = frac{f^{4} (1) }{4!} + frac{f^{5} (1) }{5!} (x-1) + frac{f^{6} (1) }{6!} (x-1)^2 + frac{f^{7} (1) }{7!} (x-1)^3$$

$$1 = frac{f^{4} (-1) }{4!} + frac{f^{5} (-1) }{5!} (x+1) + frac{f^{6} (-1) }{6!} (x+1)^2 + frac{f^{7} (-1) }{7!} (x+1)^3$$

For more context into the idea behind finding remainders, see the answer I made for the question I asked myself over here Posted on Categories Articles