Suppose we have a polynomial equation with roots 3, 5, and 8. One such polynomial equation is x³ - 16x² + 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:

a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ... + a_n-1x + a_n = 0

We can say that the sum of the roots taken 2 at a time is a₂/a₀.

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 x³.

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 = a₂/a₀. 55 = a₂/3. Therefore, a₂ = 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/a₀. If n is an even number, then the sum of the roots taken n at a time is a_n/a₀.

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 x⁴ is 2, and the coefficient of x² 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.

1.

If a cubic equation has roots 1,2, and 3, list the roots taken 2 at a time.

2.

In the example above, what is the sum of the roots taken 2 at a time?

3.

If the leading coefficient of the equation in #1 is 12, what is the coefficient of x²?

4.

If the roots of a fifth degree polynomial are taken 4 at a time, how many sets of roots are there?

5.

Two of the roots of a cubic polynomial are 5 and 2. The coefficient of x² is 34, and the leading coefficient is 2. What is the third root?

6.

One of the roots of a cubic equation is 7. The sum of the roots taken two at a time is 31, and the sum of the roots is 11. What are the other two roots?

7.

Three of the roots of a fourth degree equation are 2, 3, and -1. The sum of the roots taken three at a time is 12. What is the missing root?

8.

In the question above, what is the sum of the roots taken two at a time?

9.

In the question above, if the leading coefficient is 1, what is the polynomial?