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 r1
and r2: Equating coefficients, we see that:

a = a
b = −a(r1 + r2)
c = ar1r2

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

Similarly, for the cubic with roots r1, r2, r3, we get the following equality: And similarly for the general polynomial with roots r 1, . . . , rn, 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

## 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 equation substitute This 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

Let be a polynomial and let be a reduced rational root of P. Then and 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.
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 that for all x that keep both sides of
the equality defined. If P(1) = 1, find P(−1).

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