根据“零”书写多项式

Fun fact: the Binomial Theorem is a special case of Vieta's formulas when all of the roots are equal. Think about it, try expanding $(x-a)^n$! Use combinatorics to compute the $i$th symmetric sum.

The Binomial theorem is a theorem which is used to expand powers of a binomial expression (indicated as $(x+y)^n$). For example: $(x+y)^2 = x^2+2xy+y^2$. Notice that for each individual term, the sum of the powers of the variables is always equal to $2$.

In general,

$(x+y)^n = \binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+...+\binom{n}{n-1}x^1y^{n-1}+\binom{n}{n}x^0y^n$

Notice that the coefficients of the terms in this expansion are the numbers in a row in Pascal's triangle where $n$ is the row number (Note: Pascal's triangle starts with row $0$).

Practice Problem: What is the coefficient of the fifth term in the expansion of $(2x+3y)^{10}$?

The first term is $\binom{10}{0}(2x)^{10}(3y)^0$, the second term is $\binom{10}{1}(2x)^{9}(3y)^1$,...the fifth term should be $\binom{10}{4}(2x)^{6}(3y)^4$. So ignoring the variables, the coefficient is $\binom{10}{4}2^63^4=1088640$.

Vieta's formula(s) are used in polynomials when trying to solve for the roots. It relates an expression of the roots with the coefficients of the polynomial.

Quadratic:

Let's say $r_1$ and $r_2$ are the roots of a quadratic, we can write out the quadratic:

$a(x-r_1)(x-r_2) = ax^2-a(r_1+r_2)x+a(r_1r_2)$

This equation is most often written as $ax^2+bx+c$. We can see that since $-a(r_1+r_2)x=bx$, $r_1+r_2=-\frac{b}{a}$. For the product of the roots, we see that $r_1r_2=\frac{c}{a}$. (This method of equating coefficients is sometimes called the method of undetermined coefficients.)

Let's try an example:

What is the sum of the roots and product of the roots in the quadratic, $x^2+7x+10$?

Well, the sum of the roots is just $-\frac{b}{a}$ which is $-\frac{7}{1} = -7$. The product of the roots is just $\frac{c}{a}$ which is $\frac{10}{1} = 10$.

Vieta's can be extended to higher degree polynomials, not just quadratics!

General Formula:

Given a polynomial where the roots are {$x_1, x_2, ..., x_n$}:

$f(x) = a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x^1+a_0$

The sum of all the roots:

$x_1+x_2+x_3+x_4+...+x_n=-\frac{a_{n-1}}{a_n}$

The sum of all the combinations of products between two roots:

$x_1x_2+x_1+x_3+...x_{n-1}x_n=\frac{a_{n-2}}{a_n}$

The sum of all the combinations of products between three roots:

$x_1x_2x_3+x_1x_2x_4+...x_{n-2}x_{n-1}x_n=-\frac{a_{n-3}}{a_n}$

...

The product of all the roots:

$x_1x_2x_3...x_{n-1}x_n = (-1)^n\frac{a_0}{a_n}$

More information on Vieta's Formulas can be found [here].(http://artofproblemsolving.com/wiki/index.php/Vieta's_Formulas)

Vieta's Formulas

Introduction through Observation

Let $f(x)=x^2+5x+6$.

You may have noticed that the sum of the roots was $-b=-5$. You may also have noticed that the product of the roots is $c=6$. Lets try to prove our suspicions. For now, we will start with quadratics with a leading coefficient of $1$.

Let $r$ and $s$ be the roots of the quadratic.

Expanding $(x-r)(x-s)$ yields $$x^2-sx-rx+rs.$$ Simplifying further yields $$(x-r)(x-s)=x^2-(r+s)x+rs.$$ Aha! Our suspicions were indeed correct! We notice that $b=-(r+s)$ which means that the sum of the roots is $-b$. We also see that the product of the roots is $c$.

Warning: If the constant term of the quadratic is negative, the product will be negative as well.

However, we are not done. We only covered what happens for quadratics with a leading coefficient of one. Lets see what happens when we alter the value of $a$.

Again let our roots be $r$ and $s$. But this time, let $a\neq 1$. Our factored form is $$a(x-r)(x-s)=0.$$ Expanding yields $$\begin{align} &a(x-r)(x-s) \\ &=a(x^2-(r+s)x+rs) \\ &=ax^2-a(r+s)+ars \end{align}$$

Thus $b=-a(r+s)$ so $-b=a(r+s)$. Now in order to get our sum of $r+s$, we must divide by $a$ to yield $$-\frac{b}{a}=r+s.$$ Therefore, the sum of the roots is $-\frac{b}{a}$.

In the same way, we see that $c=ars$. To get $rs$, we must divide by $a$. Therefore, the product of the roots is $\frac{c}{a}$.

Important: The sum of the roots of any quadratic is negative the quotient of the linear term and the squared term.

Important: The product of the roots of any quadratic is the quotient of the constant term and the squared term.

Next we will take a look at cubics and beyond! (Coming soon).

Vieta's formulas for degree $n$ polynomials, forwards and backwards

I. Introduction

Named after François Viète, Vieta's formulas are formulas that relate the coefficients of a polynomial to the sums and products of its roots. These formulas are especially helpful in algebra, in which it can often solve a problem more efficiently and quicker.

II. Recap: Vieta's Formulas for a Quadratic

If we have a quadratic $ax^2+bx+c=0$ with solutions $x_1$ and $x_2$, $$\begin{cases} x_1+x_2=-\frac{b}{a}\\ x_1 x_2=\frac{c}{a} \end{cases}$$

(For a full explanation, please click here.)

III. Generalization of Vieta's Formulas

Similarly, the same procedure can be done with a polynomial $P(x)$ of degree $n$: \begin{equation} P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_2x^2+a_1x+a_0 \end{equation}

$$\begin{cases} x_1+x_2+...+x_{n-1}+x_n=-\frac{a_{n-1}}{a} \\ x_1 x_2+x_2 x_3...+x_{n-2} x_{n-1} + x_{n-1}x_n=\frac{a_{n-2}}{a} \\ x_1 x_2 x_3+x_2 x_3 x_4...+x_{n-3} x_{n-2} x_{n-1} + x_{n-2} x_{n-1} x_n=-\frac{a_{n-3}}{a} \\ ... \\ x_1 x_2 ... x_{n-1} x_n=(-1)^n \frac{a_0}{a} \end{cases}$$

Try this problem!

Let $r$, $s$, and $t$ be the roots to the polynomial $3x^3-6x^2-x+5$. Find $r^2+s^2+t^2+2 r s+2 r t+2 s t$.

Solution

You might recognize that $r^2+s^2+t^2+2 r s+2 r t+2 s t = (r+s+t)^2$. Using Vieta's formula for the sum of the roots, you get \begin{equation} (r+s+t)^2=(-\frac{-6}{3})^2=2^2=\boxed{4.} \end{equation}

$\underline{\text{Topic: Vieta's formulas for cubics, forwards and backwards}}$

I. Recap: Vieta's Formulas for a Quadratic

If we have a quadratic $ax^2+bx+c=0$ with solutions $x_1$ and $x_2$, $$\begin{cases} x_1+x_2=-\frac{b}{a}\\ x_1 x_2=\frac{c}{a} \end{cases}$$

(For a full explanation, please click here.)

II. Vieta's Formulas for a Cubic

Similarly, the same applies to a cubic polynomial. If we have a cubic \begin{equation} ax^3+bx^2+cx+d=0 \end{equation} with solutions $x_1, x_2$, and $x_3$, $$\begin{cases} x_1+x_2+x_3=-\frac{b}{a}\\ x_1 x_2 + x_1 x_3 + x_2 x_3=\frac{c}{a} \\ x_1 x_2 x_3=-\frac{d}{a} \end{cases}$$

(Video) Write a Polynomial Function in Terms of Its Roots

by James Elliott

In this video by James Elliott, you will learn to write a polynomial function in standard form by its roots.

Summary

Given the four roots: $x=0, -3, \pm1$, we can write the simplest polynomial.

We will take each root, $k$, and write it in factor form, $x-k$. Therefore, we can start writing our polynomial:

$$f(x) = (x-0)(x-(-3))(x-1)(x-(-1))$$

When we clean up this polynomial, we get:

$$f(x)=x(x+3)(x-1)(x+1)$$

Now that it's cleaner, we want to write it in standard form. The first two factors are simple to multiply together. The last two make up a difference of squares. So, we will end up with:

$$f(x) = (x^2 + 3x) (x^2 -1)$$

Lastly, we can simplify and make sure it's written in standard form.

$$f(x) = x^4 + 3x^3 -x^2-3x$$

Here is the simplest polynomial function possible, which we found my multiplying the roots.

Write a Polynomial In Terms of Roots

In this lesson, we're going to learn how to write a polynomial when you are given its roots. Roots are the values of $x$ that make a polynomial equal to zero. If you know the roots, you can work backwards to build the polynomial.

Let's say you're given the roots $2$, $-3$, and $5$. To write the polynomial, first turn each root into a factor. That means you write:

$$(x-2), (x+3), (x-5)$$

Watch your signs!

A polynomial is made from multiplying all factors together. So, you can write:

$$(x-2)(x+3)(x-5)$$

Now, we multiply the factors step by step. First, multiply the first two:

$$(x-2)(x+3) = x^2 + x -6$$

Next, multiply that result by the last factor:

$$(x^2+x-6)(x-5)$$

Do the multiplication carefully (you can use distribution or box method), and you'll get:

$$x^3-4x^2-11x+30$$

That’s your polynomial!

If you had a complex root like $2+ i$, remember from the Conjugate Root Theorem that $2 - i$ must also be a root. Always include the conjugate!