# Roots Taken N at a Time

Reference > Mathematics > Algebra > PolynomialsSuppose we have a polynomial equation with roots 3, 5, and 8. One such polynomial equation is x^{3} - 16x^{2} + 79x - 120 = 0.

In that polynomial, 16 is the sum of the roots, and 120 is the product, but what about that 79? Where does that come from? It's the sum of the roots taken 2 at a time.

We can find the sum of the roots taken two at a time as follows:

First, we find every possible way of pairing the roots in groups of two:

3 and 5

5 and 8

3 and 8

Now we multiply those pairs together:

3 x 5 = 15

5 x 8 = 40

3 x 8 = 24

And now we add those products together:

15 + 40 + 24 = 79.

If we use the way of writing polynomial equations that we used in the last section:

a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2} + ... + a_{n-1}x + a_{n} = 0

We can say that the sum of the roots taken 2 at a time is a_{2}/a_{0}.

**Example:** In a 5^{th} degree polynomial, the leading coefficient is 3, and the roots are 1, 2, 2, 3, and 4. Find the coefficient of x^{3}.

**Solution:** The products of the roots taken two at a time are 1 x 2 = 2, 1 x 2 = 2, 1 x 3 = 3, 1 x 4 = 4, 2 x 2 = 4, 2 x 3 = 6, 2 x 4 = 8, 2 x 3 = 6, 2 x 4 = 8, and 3 x 4 = 12. The sum of these is 55. Thus, 55 = a_{2}/a_{0}. 55 = a_{2}/3. Therefore, a_{2} = 165.

We can extend this idea as follows:

If n is an odd number, then the sum of the roots taken n at a time is -a_{n}/a_{0}. If n is an even number, then the sum of the roots taken n at a time is a_{n}/a_{0}.

**Example:** Find the sum of the roots taken 3 at a time, if the roots are 1,4,2, and 1.

**Solution:** The combination of roots 3 at a time are: 1x4x2 = 8, 1x4x1 = 4, 1x2x1 = 2, and 4x2x1 = 8. Thus, the sum of the roots taken 3 at a time is 22.

**Example:** In a fourth degree equation, the roots are 3, 4, 5, and x. The coefficient of x^{4} is 2, and the coefficient of x^{2} is 402. Find the missing root.

**Solution:** The sum of the roots taken 3 at a time is 3(4)(5) + 3(4)x + 3(5)x + 4(5)x = 60 + 12x + 15x + 20x = 60 + 47x = 402/2. Therefore, x = 3.

**Example:** Find the coefficient of x in a 5th degree polynomial equation with no leading coefficient, if the roots are 1, 1, 2, 1, and 1/2.

**Solution:** The sum of the roots taken 4 at a time is 1x1x2x1 + 1x1x2x(1/2) + 1x1x1x(1/2) + 1x2x1x(1/2) + 1x2x1x(1/2) = 2 + 1 + (1/2) +1 + 1 = 11/2.

## Questions

^{2}?

^{2}is 34, and the leading coefficient is 2. What is the third root?