Question

What is the relationship between the roots and coefficients of a polynomial?

Answer 1

Manifold...

Explanation:

There are many things we can say about the relationship between the roots of a polynomial and its coefficients. Here are just a few...

Rational roots theorem

Given a polynomial:

`a_n x^n + a_(n-1) x_n-1 + ... + a_1 x + a_0" "`

with integer coefficients and `a_n != 0`

any rational zeros are expressible in the form `p/q` for integers `p, q` with `p` a divisor (positive or negative) of the constant term `a_0` and `q` a divisor of the coefficient `a_n` of the leading term.

The same is true if we specify Gaussian integers instead of integers, or indeed the elements of any integral domain.

Descartes' Rule of Signs

For any polynomial with real coefficients written in standard form, the number of changes in the signs of the coefficients gives the maximum number of positive real zeros. If the number of zeros is less, then it is less by a multiple of `2`

For example:

`x^4+x^3-4x^2+2x-5`

has coefficients with signs `+ + - + -`. With `3` changes, we can deduce that this quartic has `3` or `1` positive real zeros.

Furthermore, if you invert the signs on the terms of odd degree, then the number of changes in the resulting pattern indicates the number of negative real zeros. In our example, we would get `+ - - - -`, so with one change of sign we could tell that the quartic has exactly one negative real zero.

Discriminant

You are almost certainly familiar with the formula for the discriminant `Delta` of a quadratic `ax^2+bx+c`, namely:

`Delta = b^2-4ac`

Assuming that `a, b, c` are rational, we can interpret its value as follows:

• If `Delta > 0` is a perfect square, then the quadratic has two distinct rational zeros.

• If `Delta > 0` is not a perfect square, then the quadratic has two distinct irrational zeros.

• If `Delta = 0` then the quadratic has one repeated rational zero.

• If `Delta < 0` then the quadratic has two distinct non-real complex zeros which are complex conjugates of one another.

Polynomials of higher degree also have discriminants. They are not particularly useful for quartics and above, except to identify when they have repeated zeros, but for cubics they are very useful.

The discriminant of `ax^3+bx^2+cx+d` is given by:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

• If `Delta > 0` the cubic has `3` real zeros.

• If `Delta = 0` the cubic has a repeated zero and all three zeros are real.

• If `Delta < 0` the cubic has one real zero and two complex conjugate non-real zeros.

Answer 2

A few more...

Explanation:

Elementary symmetric polynomials

The coefficients of a monic polynomial are (modulo alternating signs), the elementary symmetric polynomials in the zeros.

For example:

`(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta`

`(x-alpha)(x-beta)(x-gamma) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma`

etc.

In particular, given a polynomial:

`a_n x_n + a_(n-1) x_(n-1) + ... + a_1 x + a_0`

the sum of its zeros is `-a_(n-1)/a_n` and the product of its zeros is `(-1)^n a_0/a_n`.

Since any symmetric polynomial can be constructed from the elementary symmetric polynomials, we can find a polynomial that has zeros that are (say) the squares of the zeros of another polynomial - without finding what the zeros actually are.

For a substantial and important application of this, see https://socratic.org/s/aPGxwybx

Coefficient sum shortcuts

If the sum of the coefficients of a polynomial is `0`, then you can infer that `1` is a zero.

If inverting the signs on terms of odd degree results in coefficients that sum to `0`, then you can infer that `-1` is a zero.

Reversing the order and reciprocals

Given a polynomial:

`a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0`

with `a_n != 0` and `a_0 != 0`

Then the polynomial:

`a_0 x_n + a_1 x_(n-1) + ... + a_(n-1) x + a_n`

has zeros which are reciprocals of the original polynomial.

To see why that is so, note that:

`1/x_n (a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0)`

`=a_n + a_(n-1) 1/x + ... + a_1 1/x^(n-1) + a_0 1/x^n`

Symmetry and reciprocals

From the preceding property, we can deduce that:

If the coefficients of a polynomial are symmetrical then you can infer that the reciprocal of any zero is also a zero.

For example:

`6x^4+5x^3-38x^2+5x+6`

has zeros `2, 1/2, -3, -1/3`