AMC 10 Daily Practice - Polynomial Functions

must have four roots, three of which are roots of . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of and are the same, we know that

where is the fourth root of . Substituting and expanding, we find that

. Comparing coefficients with , we see that

,

,

,

.

(Solution . picks up here.)

Let's solve for , , , and . Since , , so . Since , , and . Thus, we know that

.

Taking , we find that

,

,

.

A faster ending to Solution is as follows. We shall solve for only a and .

Since , , and since , .Then,

,

,

.

We notice that the constant term of and the constant term in . Because can be factored as (where r is the unshared root of , we see that using the constant

Now we once again write out in factored form:

.

We can expand the expression on the righthand side to get:

Now we have

.

Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations

, and finally,

.

We know that is the sum of its coefficients, hence . We substitute the values we obtained for and into this expression to get .

Let , , and be the roots of , Let be the additional root of .Then from Vieta's formulas on the quadratic term of and the cubic term of , we obtain the following:

,

,

Thus .

Now applying Vieta's formulas on the constant term of , the linear term of , and the linear term of , we obtain:

.

.

.

Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:

.

It follows that . But so .

Now we can factor in terms of as .

Then and .

Hence .

Let the roots of be , , and . Let the roots of be , , , and . From Vieta's formula, we have:

，

．

Thus: . Since , , and are common roots, we have:

.

Then plug in for , we have:

.

.

Let ,

Note that

This gives us a pretty good guess of .

To verify, we elimiate the term by finding .This polynomial consists of the difference of two polynomials with three common factors, so it must also have these factors. The polynomial is , and must be equal to .

Equating the coefficients, we get three equations. We will tackle the situation one equation at a time, starting with the tems. Looking at the coefficients, we get . .The solution to the previous is obviously . We can now find and .

∴ and . Finally, ,

∴

Solving the original problem ..