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### Sums and Products of Roots of Polynomials

```Date: 07/24/2002 at 07:09:19
From: Maryanne
Subject: Sum and product of the roots of cubic and quartic equations

In a quadratic equation, the sum of the roots is -b/a and the product
of the roots is c/a.  I know there are formulas for the sum of the
roots of a cubic and quartic equation taken one at a time, taken two
at a time...   And the formulas for the product of the roots of a
cubic and quartic equation, I also would like to know.

Thanks.
```

```
Date: 07/24/2002 at 16:56:21
From: Doctor Greenie
Subject: Re: Sum and product of the roots of cubic and quartic
equations

I don't see a full answer to your question in the Dr. Math archives,
so I'm going to write a lengthy response to your question.

Let's first look at the relatively familiar fact that

given quadratic equation  ax^2 + bx + c = 0

(i)  the sum of the roots is (-b/a); and
(ii) the product of the roots is (c/a)

Let's see why this is so....

Suppose we represent the roots by r1 and r2.  (I am using a notation
here in which the number following the "r" is a subscript -- "r1"
and "r2" are read as "r one" and "r two", or as "r sub one" and "r
sub two".)  Then an equation with those roots is

(x-r1)(x-r2) = 0

and the specific equation with leading coefficient "a" is

a(x-r1)(x-r2) = 0

FOILing out the left side of this equation, we get

a(x^2 - r1x - r2x + r1r2) = 0

a(x^2 - (r1+r2)x + r1r2) = 0

ax^2 + (-a(r1+r2))x + a(r1r2) = 0

Now we compare this to our general form of a quadratic equation

ax^2 + bx + c = 0

and we see that

(1) b = -a(r1+r2)

From this, we can see algebraically that the sum of the roots,
(r1+r2), is equal to (-b/a).  We also see that

(2) c = a(r1r2)

From this, we can see algebraically the the product of the roots,
(r1r2), is equal to (c/a).

Now let's do the same type of analysis for a cubic (third degree)
equation.  Suppose the roots are r1, r2, and r3; the equation with
leading coefficient a with these roots is

0 = a(x-r1)(x-r2)(x-r3)

= a(x-r1)(x^2 - r2x - r3x + r2r3)

= a(x-r1)(x^2 - (r2+r3)x + r2r3)

= a(x^3 - (r2+r3)x^2 + r2r3x - r1x^2 + r1(r2+r3)x - r1r2r3)

= a(x^3 - (r1+r2+r3)x^2 + (r1r2 + r1r3 + r2r3)x - r1r2r3)

= ax^3 - a(r1+r2+r3)x^2 + a(r1r2 + r1r3 + r2r3)x - a(r1r2r3)

If we compare this final form of this equation to a cubic equation in
the "standard" form

ax^3 + bx^2 + cx + d = 0

we see that

(1) b = -a(r1+r2+r3)

and so again the sum of the roots (r1+r2+r3) is equal to -b/a,
where "b" is the coefficient of the term one degree less than the
degree of the polynomial.

We also see that

(2) d = -a(r1r2r3)

and so now we have that the product of the roots (r1r2r3) is equal
to -d/a, where "d" is the constant term of the polynomial.

Note here that this is different from the results for a quadratic
equation:  for the quadratic equation, the product of the roots was

(constant term)
---------------------

whereas for the cubic equation the product of the roots is

-(constant term)
---------------------

Comparing the analyses for the second- and third-degree polynomials,
you can see that the change in sign is because of the odd and even
degrees of the polynomials.

Note we also see for a third-degree polynomial that

(3) c = a(r1r2+r1r3+r2r3)

If we call "r1r2+r1r3+r2r3" the "sum of the roots taken two at a
time", then we have that this sum is c/a.

This sum of the roots taken two at a time may at first seem of little
use; however, i have seen numerous questions on tests from math
competitions which ask you to find an equation if the sum of the
reciprocals of the roots is some given number.  If the roots are r1,
r2, and r3, then the sum of the reciprocals of the roots is

1    1    1   r2r3+r1r3+r1r2
-- + -- + -- = --------------
r1   r2   r3       r1r2r3

This fraction has the sum of the roots taken two at a time as the
numerator and the product of the roots as the denominator, so knowing
that the sum of the roots two at a time is c/a is useful.

Summarizing the results for a polynomial equation of degree 3:

... sum of roots taken one at a time = -b/a

... sum of roots taken two at a time = c/a

... sum of roots taken three at a time = -d/a

And of course for a polynomial of degree 3, the "sum of the roots
taken three at a time" is just the product of the roots.

You can go through the same analysis for your fourth-degree
equation.  By comparing the equations

a(x-r1)(x-r2)(x-r3)(x-r4) = 0

and

ax^4 + bx^3 + cx^2 + dx + e = 0

you will find that

... sum of roots taken one at a time = -b/a

... sum of roots taken two at a time = c/a

... sum of roots taken three at a time = -d/a

... sum of roots taken four at a time = e/a

And similar analyses for higher-degree polynomials will show similar
results.  For example, for the polynomial

ax^10 + bx^9 + cx^8 + ... + ix^2 + jx + k

you can determine that

... sum of roots taken one at a time = -b/a

... sum of roots taken two at a time = c/a

... sum of roots taken three at a time = -d/a

...  (etc., etc.,...)

... sum of roots taken eight at a time = i/a

... sum of roots taken nine at a time = -j/a

... sum of roots taken ten at a time = k/a

Summarizing the major results of these analyses, we have the
following, for the roots of a polynomial of degree n:

(1) the sum of the roots is always (-b/a), where a and b are the
coefficients of the x^n and x^(n-1) terms of the polynomial; and

(2) if we denote the constant term of the polynomial by "C(0)", then
the product of the roots ("the sum of the roots taken n at a time")
is equal to (C(0)/a) if n is even or to (-C(0)/a) if n is odd.

Of less frequent usefulness is that, if we denote the coefficient of
the linear term of the polynomial by "C(1)", then the sum of the
roots taken (n-1) at a time is (-C(1)/a) if n is even or to (C(1)/a)
if n is odd.

I hope all of this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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