# Advanced Lecture Polynomials

Most algebra problems become straightforward to one armed with the following toolkit:

## 1 Vieta’s Formulas

If you think you haven’t seen Vieta(aka Viete)’s formulas in action , think
again. These

formulas are trivial to derive but are among the most useful tools for tackling
standard

polynomial problems. To understand Vieta, simply equate a polynomial to the
product of

its linear factors and a constant . To start, let’s consider the tame quadratic
with roots r_{1}

and r_{2}:

Equating coefficients, we see that:

a = a

b = −a(r_{1} + r_{2})

c = ar_{1}r_{2}

These are Vieta’s formulas, unsimplified. That’s all there is to it.

Similarly, for the cubic with roots r_{1}, r_{2}, r_{3}, we get the following
equality:

And similarly for the general polynomial with roots r _{1}, . . . , r_{n}, we get
that

Specifically, the product of the roots is the coefficient of the constant
term divided by the

coefficient of the leading term times −1 if the polynomial has odd degree. The
sum of the

roots is minus the coefficient of the second term from the left divided by the
coefficient of

the leading term.

## 2 Finding Roots

A stereotypical algebra problem asks you to solve an equation . Luckily, for
polynomials this

is relatively easy. For quadratics and cubics there is a straightforward method.
For quartics

and above, you will usually have to employ a trick such as a substitution or a
geometric

interpretation.

### 2.1 Quadratic and Depressed Cubic Formulae

You should know that:

To depress the cubic equationsubstituteThis
will

eliminate the x ^2 term. Now divide by a and we get a cubic of the form x^3 + bx
+ c = 0. One

of the roots of this cubic is

### 2.2 Rational Root Theorem

Letbe
a polynomial and letbe
a reduced rational root of P. Thenand

To prove this, consider the factorization of P(x).

## 3 Factoring Polynomials

In this section, completely factoring a polynomial will be defined as
expressing it as a product

of polynomials with integral coefficients that are all irreducible across the
integers by hand.

(No TI-89 factoring) Sometimes solving a problem comes down to your ability to
factor a

polynomial. True, factoring often comes down to guesswork, but then again, so do
differential

equations, and that’s a whole field in itself. You should view a factoring
exercise as a puzzle:

each term you factor out brings you closer towards the answer. Usually, there
will be a key

step of this puzzle , which is represented by a big factor. Sometimes you may
even have to

factor functions other than polynomials, but polynomials are typically the
hardest functions

to factor.

### 3.1 Tips and Trix

1. Always start by trying to factor out small things and make the resulting
polynomial

easier to work with. There’s almost always a linear factor . Difference of
squares is very

common.

2. Remember that (ax − b) is a factor of P(x) iff P( b/a ) = 0.

3. Roots of unity / cyclotomic polynomials are irreducible (if you don’t know
about this,

you should learn it, but it won’t be covered today)

4. Based on the problem’s context, is there something that should be a factor?

5. If your intuition says that the polynomial irreducible, then it probably is,
but don’t

forget about USAMO 2007/5(a one step factoring problem which nobody solved)!

### 3.2 Some polynomials to factor

Some of these might be irreducible. Hopefully these exercises will help
develop your factoring

intuition. Factor each of the following polynomials completely.

## 4 Problems

1. How many quintics are there whose roots are all 0 or 1?

2. Find all the real roots of x^3 + x + 2 = 0.

3. Let x, y, z be real numbers which add up to 0. Find all possible values of

4. Find the sum of the squares of the roots of x^4 + 2x^3 − 3x^2 + 4x − 5 = 0.

5. Find all the real roots of x^4 + (2 − x)^4 = 34.

6. If x is real and x^3 + 4x − 8 = 0, find all possible values of x^7 + 64x^2.

7. Find the sum of the absolute values of the roots of x^4 − 4x^3 − 4x^2 + 16x − 8 = 0.

8. Suppose P(x) is a polynomial such thatfor
all x that keep both sides of

the equality defined. If P(1) = 1, find P(−1).

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