# Sum and product of roots of a polynomial

Let's say we had a $$n$$th degree polynomial equation $$a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0=0$$, with $$a$$ being real coefficient. What would the sum and product of its roots be(in terms of $$a$$)? I think I got the product one but not the sum.

For the product:

Let's say the polynomial's roots are $$r_1,r_2,r_3,\ldots,r_n$$.

Then polynomial can be factored as:

$$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$$

We can set this equal to original polynomial:

$$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0=0$$

Compare constant terms:

$$a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_2x^2+a_1x+a_0$$ constant term=$$a_0$$.

$$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$$ constant term=$$(-1)^n*(\frac{r_1}{a_n})*r_2*r_3*\cdots*r_n$$

$$a_0=(-1)^n*(\frac{r_1}{a_n})*r_2*r_3*\cdots*r_n$$

Multiply $$(-1)^na_n$$ both sides:

$$r_1*r_2*r_3*\cdots r_n=(-1)^na_0a_n$$

Is this correct? Also, what can I do for the sum of the roots(I think we use the coefficients of $$x^{n-1}$$)?

EDIT: J.W. Tanner has noted in his comment that this is Vieta's Formulas which is exactly what I was looking for but couldn't find.

• $a_n(x-\frac{r_1}{an})$ should be $a_n(x-r_1)$ – Alexey Burdin Jun 28 at 20:06
• – J. W. Tanner Jun 28 at 20:13
• @J.W.Tanner Oh thanks a lot! This is exactly what I am looking for. – Aiden Chow Jun 28 at 20:15

$$a_n$$ is not a root, it is the leading coefficient.

Imagine the polynomial $$(x-2)(x-3) = x^2 - 5x + 6$$.
The leading coefficient is $$1$$ but the roots are $$2$$ and $$3$$.

This part of your argument is incorrect.

Then polynomial can be factored as:

$$a_n(x-\frac{r_1}{a_n})(x-r_2)(x-r_3)\ldots(x-r_n)$$

Then polynomial can be factored as:

$$a_n(x-{r_1})(x-r_2)(x-r_3)\ldots(x-r_n)$$

From this representation it gets obvious that the free term is

$$a_n(-1)^nr_1 r_2 ... r_n$$

But on the other hand we know it is $$a_0$$.

So the product of the roots must be:

$$\frac{a_0}{(-1)^na_n} = (-1)^n \cdot \frac{a_0}{a_n}$$

For calculating the sum of the roots just compare the coefficient before $$x^{n-1}$$.
You will get that the sum of the roots equals $$-\frac{a_{n-1}}{a_n}$$