# Vieta s Formulas and the Identity Theorem

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1 Vieta s Formulas and the Identity Theorem This worksheet will work through the material from our class on 3/21/2013 with some examples that should help you with the homework The topic of our discussion will be polynomials The first thing we ll do is define a polynomial: Definition: A polynomial is an expression of the form P (x) = x n + 1 x n x n a 1 x 1 + a 0 where x is a variable (you can plug in anything you want for x) and the a 0, a 1, a 2, are all constants We call these constants the coefficients of the polynomial We call the highest exponent of x the degree of the polynomial This is a polynomial: P (x) = 5x 3 + 4x 2 2x + 1 The highest exponent of x is 3, so the degree is 3 P (x) has coefficients = 5 a 2 = 4 a 1 = 2 a 0 = 1 Since x is a variable, I can evaluate the polynomial for some values of x All this means is that I can plug in different values for x so that my polynomial simplifies to a single number: P (1) = 5 (1) (1) 2 2 (1) + 1 = 8 P (0) = 5 (0) (0) 2 2 (0) + 1 = 1 P ( 1) = 5 ( 1) ( 1) 2 2 ( 1) + 1 = 2 Practice Problem: For practice, let s analyze the polynomial P (x) = x 3 + x 2 Write down all the coefficients of P (x), find the degree of P (x), and evaluate P ( 1), P (2), and P (1) Solution: a0 = 2, a1 = 1, a2 = 0, a3 = 1, P ( 1) = 4, P (2) = 8, P (1) = 0 The degree is 3 Notice that in the practice problem above, P (1) = 0 Values at which complicated polynomials simplify to zero are very important and are known as the roots of the polynomial 1

2 Definition: The roots of the polynomial P (x) = x n + 1 x n x n a 1 x 1 + a 0 are the values r 1, r 2, r 3, such that The roots of the polynomial are P (r 1 ) = 0 P (r 2 ) = 0 P (r 3 ) = 0 P (x) = x 3 + 7x 2 + 6x r 1 = 1 r 2 = 6 r 3 = 0 You should check that P ( 1) = 0, P ( 6) = 0, and that P (0) = 0 Practice Problem: Find two roots of the polynomial P (x) = x 2 1 (Try guessing some small numbers) What is the degree of P (x)? Solution: r1 = 1, r2 = 1 The degree is 2 Perhaps you have noticed that in the last two examples the number of roots is the same as the degree of the polynomial This is not just a coincidence - there is a theorem that says that this will always be true: Theorem 1: A polynomial of degree n has exactly n roots However, some of the roots may be very complicated (some may be complex numbers) Don t be discouraged if you cannot immediately find all n roots The polynomial P (x) = 5x 3 + 4x 2 2x + 1 has degree 3 (n = 3), so it has exactly three roots: r 1, r 2, and r 3 However, finding these roots would be really hard (in fact, two of them are complex and the other one is very messy!) Although many polynomials have very complicated roots, we can often determine a lot about them using Vieta s Formulas, the first of which we look at next: Theorem 2: Given a polynomial P (x) = x n + 1 x n x n 2 + +a 1 x 1 +a 0, with roots r 1, r 2, r 3, r n, the sum of the roots is given by r 1 + r 2 + r r n = 1 2

3 Let s look at the polynomial P (x) = 2x 4 6x 3 + 5x 2 + x 3 Notice that this polynomial has degree 4 (so n = 4), so by Theorem 1 we know that it has exactly 4 roots: call them r 1, r 2, r 3, r 4 Without knowing anything else about P (x), Theorem 2 gives us a very easy way to find the sum of the roots: r 1 + r 2 + r 3 + r 4 = a 4 = 6 2 = 3 Practice: Find the sum of the roots of the polynomials P (x) = 3x 3 + 2x 2 6x 3 and G(x) = x 2 1 Solution: P (x) has degree 3 (n = 3), so the sum of the three roots of P (x) is a2 = 2 a3 3 G(x) has degree 2 (n = 2), so the sum of the three roots of G(x) is a1 = 0 a2 1 = 0 Hopefully you find it really cool that just by looking at a polynomial you can determine the sum of its roots But we re not done quite yet Vieta s Formulas are a set of n equations, where sum of the roots is just the first one Theorem 3 (Vieta s Formulas): Consider the polynomial P (x) = x n + 1 x n x n a 1 x 1 + a 0 with degree n By Theorem 1 P (x) has n roots, call them r 1, r 2, r n Vieta s Formulas say that r 1 + r 2 + r r n = 1 (r 1 r 2 + r 1 r 3 + r 1 r 4 + r 1 r n ) + (r 2 r 3 + r 2 r r 2 r n ) + + r n 1 r n = 2 (r 1 r 2 r 3 + r 1 r 2 r 4 + r 1 r 2 r 5 + r 1 r 2 r n ) + (r 1 r 3 r 4 + r 1 r 3 r r 1 r 3 r n ) + + r n 2 r n 1 r n = 3 r 1 r 2 r 3 r 4 r n = ( 1) n a 0 These formulas look quite complicated - below you ll find a simplified version of them But before before you rush to see the Simplified Vieta s Formulas, let s notice some patterns The first equation we are already comfortable with: all it says is that the sum of the roots is an 1 Now let s look at the second equation The left hand side is a sum of every possible product of a pair of roots The first parentheses has every possible pair that involves r 1, the second has every possible pair that involves r 2, etc Now let s move on to the third equation Now the left hand side is a sum of every possible product of three roots (the parentheses are organized similarly) Hopefully the pattern is clear now, so it makes sense that the last equation is a product of all the roots (every possible sum of n roots) The right hand side is a little bit easier: it is always of the form ± a #, where the sign (±) alternates with every equation If the degree (n) is even, then the last equation ends with a positive sign and if n is odd then it ends on egative sign That s why we have the ( 1) n in the last equation Now let s look at a simpler version of Vieta s Formulas 3

4 Simplified Vieta s Formulas: In the case of a polynomial with degree 3, Vieta s Formulas become very simple Given a polynomial P (x) = x 3 + a 2 x 2 + a 1 x + a 0 with roots r 1, r 2, r 3, Vieta s formulas are r 1 + r 2 + r 3 = a 2 r 1 r 2 + r 1 r 3 + r 2 r 3 = a 1 r 1 r 2 r 3 = a 0 Find the sum of the roots and the product of the roots of the polynomial P (x) = 3x 3 + 2x 2 x + 5 By the first of the Simplified Vieta s Formulas, the sum of the roots is and the product is r 1 + r 2 + r 3 = a 2 = 2 3 r 1 r 2 r 3 = a 0 = 5 3 Practice: Find the roots of the polynomial P (x) = x 3 x using Vieta s Formulas (Hint: first figure out what the product of the roots must be Then, figure out what the sum of the roots must be Then you should be able to guess what the roots are) Solution: Applying the Simplified Vieta s Formulas we find that the sum of the roots and the product of the roots must be 0 Since the product of the roots is zero, at least one of the roots must be zero Since the sum of the roots is zero, the other two roots must be negatives of each other (or they are all zero) Now you can either plug all of this into the last of Vieta s Formula, or you can guess that the roots are 1, 1, and 0 The last topic that we covered (very quickly) was the Identity Theorem This theorem tells us when two polynomials are actually the same polynomial Theorem 4 (Identity Theorem): Suppose we have two polynomials: P (x) and G(x), and both of them have degree less than n Suppose there are n values α 1, α 2, α n such that P (α 1 ) = G(α 1 ) P (α 2 ) = G(α 2 ) P (α n ) = G(α n ) (This just means that P (x) and G(x) evaluate to the same thing for n different values of x) Then P (x) and G(x) are the same polynomial 4

5 Suppose I give you two polynomials: P (x) = ax 2 + bx + c and G(x) = Ax 2 + Bx + C All you know about them is that P (1) = G(1), P (5) = G(5), and that P ( 1) = G( 1) Then you can guarantee that P (x) and G(x) are actually the same polynomial (so a = A, b = B, and c = C) Indeed, since both P (x) and G(x) have degree less than 3 (both have degree two), the Identity Theorem tells us that if P and G evaluate to the same thing at three values of x then they are the same polynomial Well, I told you that they evaluate to the same thing at 1, 2, and 5, so they must be the same polynomial Practice: You are given a polynomial P (x) that has degree less than 10 Suppose that P (x) has at least 10 roots (ie, P (x) = 0 for at least 10 different values of x) Prove that P (x) is always zero Solution: Define the polynomial G(x) so that it is zero everywhere We can write this as G(x) = 0 (so all of the coefficients are 0, and the degree is 0) I will use the Identity Theorem to prove that G(x) is the same polynomial as P (x), which would mean that P (x) is zero everywhere (always zero) Notice that P (x) is zero at at least 10 values of x, so P (x) evaluates to the same thing as G(x) at these 10 values of x (because G(x) is always zero) Since the degrees of P (x) and G(x) are both less than 10, by the Identity Theorem this means that P (x) and G(x) are the same polynomial Therefore P (x) is everywhere zero That s it! Congratulations on reaching the end! Shoot me an to let me know that you ve gone through this worksheet and I ll bring you some candy next week! 5

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