Find the roots of a 6th order Taylor polynomial

Question

Let $P(x)$ denote the sixth-order Taylor polynomial of $$e^{-2x}-3x\cos x+5\sin x$$ at $x=0$. Let $a_1,a_2,a_3,a_4,a_5,a_6$ denote the six roots (complex roots are allowed) of the equation $P(x)=0$. If $a_1+a_2+a_3+a_4+a_5+a_6=\frac mn$ where $m$ and $n$ are two positive integers with no common factors, find the exact value of $m+n$.

In the question above, I've tried expanding the Taylor series of the given function to get the 6th order Taylor polynomial. However, I'm not sure how I should go about finding all six roots of a 6th order polynomial. This is also something that I haven't been taught in my course so far, so I'm not sure if this approach is correct in the first place. Is there some important concept I'm missing out here that doesn't require solving a 6th order polynomial?

Answer 1

Observe that if a polynomial has roots $a_1,\dots,a_6$, then it can be written as $$ c(x-a_1)(x-a_2)\cdots(x-a_6). $$ If we multiply this out, we find that the coefficient of $x^5$ is $-c(a_1+a_2+\cdots+a_6)$. Can you take it from here?

Answer 2

Hint:

The Vieta's formulas state that for a $6$'th degree polynomial

$$f(x) = a_6x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0 \tag{1}\label{eq1A}$$

where $r_i$ for $1 \le i \le 6$ are the $6$ roots, including non-real and with multiplicity, you have

$$r_1 + r_2 + r_3 + r_4 + r_5 + r_6 = -\frac{a_5}{a_6} \tag{2}\label{eq2A}$$

Answer 3

The sum of the roots of a $n$th degree polynomial $$P(x) = c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0$$ by Vieta's formulas is $-c_{n-1} / c_n$. That's because it has $n$ roots, and can be factored as $$c_n (x - r_1) (x - r_2) \cdots (x-r_n) = c_n x^n - c_n(r_1 + r_2 +\cdots + r_n) x^{n-1} + O(x^{n-2})$$