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Algebra · Chapter 13

Vieta's Formulas for Polynomials

Read the sum and product of a polynomial's roots straight off its coefficients, and answer root questions without ever finding the roots.

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13.1a0%

Quadratics: Formula, Discriminant and Vieta

In this section you will be able to solve any quadratic, tell how many real roots it has, and read the sum and product of its roots off the coefficients.

The roots of 3x2+6x−53x^2 + 6x - 5 are ugly: −3±263\frac{-3 \pm 2\sqrt{6}}{3}. Yet their cubes add up to a clean −18-18. You can get that without ever writing the roots down.

Every quadratic is a(x−r)(x−s)a(x - r)(x - s). Multiply it out: ax2−a(r+s)x+arsax^2 - a(r + s)x + ars. So the coefficients already know the sum r+sr + s and the product rsrs. That is Vieta’s formula.

The quadratic formula finds the roots themselves. The part under the square root, b2−4acb^2 - 4ac, tells you what kind of roots to expect before you do any work.

Look:

Many questions ask about r2+s2r^2 + s^2, r3+s3r^3 + s^3 or 1r+1s\frac1r + \frac1s. Rewrite each one using only r+sr + s and rsrs, then plug in.

Exploreplay first, formulas later

P(x) = x² − 2x − 3; roots −1, 3; sum of roots 2, product −3

Degree2
Lead a1
r₁r₂
P(x)=(x+1)(x−3)=x2−2x−3P(x) = (x + 1)(x - 3) = x^{2} - 2x - 3
r1=−1,  r2=3r_1 = -1,\; r_2 = 3
sumr1+r2=−ba=−−21=2r_1 + r_2 = -\frac{b}{a} = -\frac{-2}{1} = 2
productr1r2=ca=−31=−3r_1r_2 = \frac{c}{a} = \frac{-3}{1} = -3
discriminantb2−4ac=(−2)2−4(1)(−3)=16b^2 - 4ac = (-2)^2 - 4(1)(-3) = 162 different real roots
vertex(h,k)=(1,−4),  P(x)=(x−1)2−4(h, k) = (1, -4),\; P(x) = (x - 1)^2 - 4

P(x) = x² − 2x − 3; roots −1, 3; sum 2; product −3; discriminant 16: 2 different real roots

Name it

Theorem 13.1.1 · Quadratic Formula

The solutions to the quadratic equation

ax2+bx+c=0ax^2 + bx + c = 0

are

x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Concept 13.1.2 · Discriminant

In the quadratic formula,

b2−4acb^2 - 4ac

(the part inside the square root) is the discriminant of the quadratic.

  1. If the discriminant b2−4acb^2 - 4ac is 00, then the quadratic has a double (repeated) root.
  2. If the discriminant b2−4acb^2 - 4ac is positive, the quadratic has 2 different real roots.
  3. If the discriminant b2−4acb^2 - 4ac is negative, the quadratic has no real roots.

Theorem 13.1.3 · Vieta's Formula For Quadratics

In a quadratic equation

ax2+bx+c=0ax^2 + bx + c = 0

the sum of its roots is

−ba-\frac{b}{a}

and the product of its roots is

ca.\frac{c}{a}.

Watch itpredict, then reveal

Example 13.1

Omega Learn

If the roots of the quadratic 3x2+6x−53x^2 + 6x - 5 are rr and ss, find r3+s3r^3 + s^3.

Vieta roots (its total is revealed at the end of the example)

P(x)=3x2+6x−5P(x) = 3x^{2} + 6x - 5
r,s=−3±263r, s = \frac{-3 \pm 2\sqrt{6}}{3}
sumr+s=−ba=−63=−2r + s = -\frac{b}{a} = -\frac{6}{3} = -2
productrs=ca=−53=−53rs = \frac{c}{a} = \frac{-5}{3} = -\frac{5}{3}
power sum S1r+s=−2r + s = -2
power sum S2r2+s2=?r^{2} + s^{2} = ?
power sum S3r3+s3=?r^{3} + s^{3} = ?

P(x) = 3x² + 6x − 5; roots −2.633, 0.633; sum −2; product −5/3; power sums −2, ?, ?

  1. 1

    The roots are −3±263\frac{-3 \pm 2\sqrt6}{3}. Cubing those is painful. What is r+sr + s instead?

    Predict first:
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Try it

Problem 13.2.1

Find the sum and product of the roots of the quadratic 2x2−24x+172x^2 - 24x + 17.

Problem 13.2.2

AMC 12

There are two values of aa for which the equation 4x2+ax+8x+9=04x^2 + ax + 8x + 9 = 0 has only one solution for xx. What is the sum of these values of aa?

P(x) = 4x² + 8x + 9; roots −1 + 1.118i, −1 − 1.118i; sum of roots −2, product 9/4

a4
b8
c9
P(x)=4(x2+2x+94)=4x2+8x+9P(x) = 4(x^{2} + 2x + \frac{9}{4}) = 4x^{2} + 8x + 9
r1,r2=−2±5i2r_1, r_2 = \frac{-2 \pm \sqrt{5}i}{2}
sumr1+r2=−ba=−84=−2r_1 + r_2 = -\frac{b}{a} = -\frac{8}{4} = -2
productr1r2=ca=94=94r_1r_2 = \frac{c}{a} = \frac{9}{4} = \frac{9}{4}
discriminantb2−4ac=(8)2−4(4)(9)=−80b^2 - 4ac = (8)^2 - 4(4)(9) = -80no real roots: a complex pair

P(x) = 4x² + 8x + 9; roots −1 + 1.118i, −1 − 1.118i; sum −2; product 9/4; discriminant −80: no real roots: a complex pair

Problem 13.2.6

AMC 10

What is the sum of the reciprocals of the roots of the equation 20032004x+1+1x=0\frac{2003}{2004}x + 1 + \frac{1}{x} = 0?

Problem 13.2.9

AMC 10

What is the sum of all real numbers xx for which ∣x2−12x+34∣=2|x^2 - 12x + 34| = 2?

Contest cornerreal contest problems

Problem 13.2.3

AMC 10

Let aa and bb be the roots of the equation x2−mx+2=0x^2 - mx + 2 = 0. Suppose that a+1ba + \frac{1}{b} and b+1ab + \frac{1}{a} are the roots of the equation x2−px+q=0x^2 - px + q = 0. What is qq?

Problem 13.2.7

AMC 10

The quadratic equation x2+mx+nx^2 + mx + n has roots twice those of x2+px+mx^2 + px + m, and none of mm, nn, and pp is zero. What is the value of np\frac{n}{p}?

Problem 13.2.8

AMC 10

Let ff be a function for which f(x3)=x2+x+1f\left(\frac{x}{3}\right) = x^2 + x + 1. Find the sum of all values of zz for which f(3z)=7f(3z) = 7.

Problem 13.2.11

AMC 10

Suppose that aa and bb are nonzero real numbers, and that the equation x2+ax+b=0x^2 + ax + b = 0 has solutions aa and bb. Then the pair (a,b)(a, b) is

Click a choice or type its letter (A–E).

Problem 13.2.14

AMC 10

Let aa, bb, and cc be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation (x−a)(x−b)+(x−b)(x−c)=0(x - a)(x - b) + (x - b)(x - c) = 0?

Checkpoint for section 13.1a

Get 2 of 3 right to master this section. One try each, no hints.
1

What is the sum of the roots of 5x2−3x−2=05x^2 - 3x - 2 = 0?

2

For which value of kk does x2+6x+k=0x^2 + 6x + k = 0 have a double root?

3

The roots of x2−7x+3=0x^2 - 7x + 3 = 0 are rr and ss. What is r2+s2r^2 + s^2?

13.1b0%

Vieta for Higher-Degree Polynomials

In this section you will be able to use Vieta's formulas on cubics and quartics, where the signs alternate.

A cubic has three roots, so there are three Vieta numbers: the sum, the sum of the products of pairs, and the product. All three sit in the coefficients.

Expand a(x−r)(x−s)(x−t)a(x - r)(x - s)(x - t). Look at the coefficients:

  • of x2x^2: −a(r+s+t)-a(r + s + t)
  • of xx: +a(rs+st+tr)+a(rs + st + tr)
  • the constant: −a⋅rst-a \cdot rst

The signs alternate because every root comes with a minus sign.

The same holds for any degree: the coefficient of xn−kx^{n-k} is ana_n times the sum of all products of kk roots, with sign (−1)k(-1)^k.

Look:

A root also satisfies its own equation. If r3−6r−5=0r^3 - 6r - 5 = 0, then r3=6r+5r^3 = 6r + 5. That turns cubes back into Vieta sums.

Exploreplay first, formulas later

P(x) = x³ − 2x² − 5x + 6; roots −2, 1, 3; sum of roots 2, product −6

Degree3
Lead a1
rst
P(x)=(x+2)(x−1)(x−3)=x3−2x2−5x+6P(x) = (x + 2)(x - 1)(x - 3) = x^{3} - 2x^{2} - 5x + 6
P(x)=ax3+bx2+cx+d=a(x−r)⋯(x−t)P(x) = ax^3 + bx^2 + cx + d = a(x - r)\cdots(x - t)
r=−2,  s=1,  t=3r = -2,\; s = 1,\; t = 3
sumr+s+t=−ba=−−21=2r + s + t = -\frac{b}{a} = -\frac{-2}{1} = 2e1=−a2a3e_{1} = -\frac{a_{2}}{a_{3}}
sum of products of 2rs+rt+st=ca=−51=−5rs + rt + st = \frac{c}{a} = \frac{-5}{1} = -5e2=a1a3e_{2} = \frac{a_{1}}{a_{3}}
productrst=−da=−61=−6rst = -\frac{d}{a} = -\frac{6}{1} = -6e3=−a0a3e_{3} = -\frac{a_{0}}{a_{3}}

P(x) = x³ − 2x² − 5x + 6; roots −2, 1, 3; sum 2; product −6

Name it

Theorem 13.1.4 · Vieta's Formula For Higher Degree Polynomials

In a polynomial

anxn+an−1xn−1+⋯+a1x+a0=0a_n x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 = 0

with roots r1,r2,r3,…,rnr_1, r_2, r_3, \dots, r_n, the following holds:

r1+r2+r3+⋯+rn=−an−1an(the sum of all terms)r_1 + r_2 + r_3 + \dots + r_n = -\frac{a_{n-1}}{a_n} \quad \text{(the sum of all terms)}
r1r2+r1r3+⋯+rn−1rn=an−2an(the sum of all products of 2 terms)r_1r_2 + r_1r_3 + \dots + r_{n-1}r_n = \frac{a_{n-2}}{a_n} \quad \text{(the sum of all products of 2 terms)}
r1r2r3+r1r2r4+⋯+rn−2rn−1rn=−an−3an(the sum of all products of 3 terms)r_1r_2r_3 + r_1r_2r_4 + \dots + r_{n-2}r_{n-1}r_n = -\frac{a_{n-3}}{a_n} \quad \text{(the sum of all products of 3 terms)}
⋮\vdots
r1r2r3⋯rn=(−1)na0an(the product of all n terms)r_1r_2r_3\cdots r_n = (-1)^n\frac{a_0}{a_n} \quad \text{(the product of all } n \text{ terms)}

The signs alternate: a sum of products of an odd number of roots gets a minus sign, an even number gets a plus sign.

Watch itpredict, then reveal

Example 13.2

Omega Learn

If rr, ss, and tt are roots of the cubic x3−6x−5=0x^3 - 6x - 5 = 0, find r3+s3+t3r^3 + s^3 + t^3.

Vieta roots (its total is revealed at the end of the example)

P(x)=x3−6x−5P(x) = x^{3} - 6x - 5
r≈−1.791,  s=−1,  t≈2.791r \approx -1.791,\; s = -1,\; t \approx 2.791
sumr+s+t=−ba=−01=0r + s + t = -\frac{b}{a} = -\frac{0}{1} = 0
sum of products of 2rs+rt+st=ca=−61=−6rs + rt + st = \frac{c}{a} = \frac{-6}{1} = -6
productrst=−da=−−51=5rst = -\frac{d}{a} = -\frac{-5}{1} = 5
power sum S1r+s+t=0r + s + t = 0
power sum S2r2+s2+t2=?r^{2} + s^{2} + t^{2} = ?
power sum S3r3+s3+t3=?r^{3} + s^{3} + t^{3} = ?

P(x) = x³ − 6x − 5; roots −1.791, −1, 2.791; sum 0; product 5; power sums 0, ?, ?

  1. 1

    The x2x^2 term is missing. What is r+s+tr + s + t?

    Predict first:
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Example 13.3

AMC 12

The polynomial p(x)=x3+ax2+bx+cp(x) = x^3 + ax^2 + bx + c has the property that the average of its zeros, the product of its zeros, and the sum of its coefficients are all equal. The yy-intercept of the graph of y=p(x)y = p(x) is 22. What is bb?

Vieta roots (its total is revealed at the end of the example)

P(x)=x3+?x2+?x+2P(x) = x^{3} + \boxed{?}x^{2} + \boxed{?}x + 2
r≈−7.502,  s≈0.206,  t≈1.296r \approx -7.502,\; s \approx 0.206,\; t \approx 1.296
sumr+s+t=−ba=−?1=?r + s + t = -\frac{b}{a} = -\frac{?}{1} = ?
sum of products of 2rs+rt+st=ca=?1=?rs + rt + st = \frac{c}{a} = \frac{?}{1} = ?
productrst=−da=−21=−2rst = -\frac{d}{a} = -\frac{2}{1} = -2
evaluateP(1)=?P(1) = ?

P(x) = x³ + ?x² + ?x + 2; roots −7.502, 0.206, 1.296; sum ?; product −2; P(1) = ?

  1. 1

    The yy-intercept is p(0)p(0). What is cc?

    Predict first:
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Try it

Problem 13.2.10

AMC 10

The polynomial x3−ax2+bx−2010x^3 - ax^2 + bx - 2010 has three positive integer roots. What is the smallest possible value of aa?

Contest cornerreal contest problems

Problem 13.2.4

AMC 10/12

A rectangular box P\mathcal{P} has distinct edge lengths aa, bb, and cc. The sum of the lengths of all 1212 edges of P\mathcal{P} is 1313, the sum of the areas of all 66 faces of P\mathcal{P} is 112\frac{11}{2}, and the volume of P\mathcal{P} is 12\frac{1}{2}. What is the length of the longest interior diagonal connecting two vertices of P\mathcal{P}?

Problem 13.2.5

AMC 12

For how many ordered pairs (a,b)(a, b) of integers does the polynomial x3+ax2+bx+6x^3 + ax^2 + bx + 6 have 33 distinct integer roots?

P(x) = x³ − 4x² + x + 6; roots −1, 2, 3; sum of roots 4, product −6

Degree3
Lead a1
r₁r₂r₃
P(x)=(x+1)(x−2)(x−3)=x3−4x2+x+6P(x) = (x + 1)(x - 2)(x - 3) = x^{3} - 4x^{2} + x + 6
r1=−1,  r2=2,  r3=3r_1 = -1,\; r_2 = 2,\; r_3 = 3
sumr1+r2+r3=−ba=−−41=4r_1 + r_2 + r_3 = -\frac{b}{a} = -\frac{-4}{1} = 4
sum of products of 2r1r2+r1r3+r2r3=ca=11=1r_1r_2 + r_1r_3 + r_2r_3 = \frac{c}{a} = \frac{1}{1} = 1
productr1r2r3=−da=−61=−6r_1r_2r_3 = -\frac{d}{a} = -\frac{6}{1} = -6

P(x) = x³ − 4x² + x + 6; roots −1, 2, 3; sum 4; product −6

Problem 13.2.12

AIME

Suppose that the roots of x3+3x2+4x−11=0x^3 + 3x^2 + 4x - 11 = 0 are aa, bb, and cc, and that the roots of x3+rx2+sx+t=0x^3 + rx^2 + sx + t = 0 are a+ba + b, b+cb + c, and c+ac + a. Find tt.

Problem 13.2.13

AMC 10

For certain real numbers aa, bb, and cc, the polynomial

g(x)=x3+ax2+x+10g(x) = x^3 + ax^2 + x + 10

has three distinct roots, and each root of g(x)g(x) is also a root of the polynomial

f(x)=x4+x3+bx2+100x+c.f(x) = x^4 + x^3 + bx^2 + 100x + c.

What is f(1)f(1)?

Problem 13.2.15

AMC 10

All the roots of the polynomial z6−10z5+Az4+Bz3+Cz2+Dz+16z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16 are positive integers, possibly repeated. What is the value of BB?

Problem 13.2.17

AMC 12A

Let aa, bb, and cc be the roots of the polynomial x3+kx+1x^3 + kx + 1. What is the sum

a3b2+a2b3+b3c2+b2c3+c3a2+c2a3?a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3?

(Give your answer in terms of kk.)

Checkpoint for section 13.1b

Get 2 of 3 right to master this section. One try each, no hints.
1

The roots of x3−5x2+2x+7=0x^3 - 5x^2 + 2x + 7 = 0 are rr, ss, tt. What is rs+st+trrs + st + tr?

2

For the same roots of x3−5x2+2x+7=0x^3 - 5x^2 + 2x + 7 = 0, what is r2+s2+t2r^2 + s^2 + t^2?

3

What is the product of the four roots of 2x4−3x3+x−8=02x^4 - 3x^3 + x - 8 = 0?

13.2

Practice Problems

In this section you will be able to use Vieta on contest problems that hide their polynomial.

Two more from the practice set. One you’ve seen in a different form; the other is a USAMO problem that Vieta cracks in four lines.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 13.2.16

AMC 10/12

This problem also appears in the book as 13.2.15.

All the roots of polynomial z6−10z5+Az4+Bz3+Cz2+Dz+16z^6 - 10z^5 + Az^4 + Bz^3 + Cz^2 + Dz + 16 are positive integers, possibly repeated. What is the value of BB?

Problem 13.2.18

USAMOChallenge

In the polynomial x4−18x3+kx2+200x−1984=0x^4 - 18x^3 + kx^2 + 200x - 1984 = 0, the product of 22 of its roots is −32-32. Find kk.

13.3

Additional Problems

In this section you will be able to mix Vieta with substitution, factoring and polynomial division on harder AIME problems.

These are the additional problems. Most are AIME: the polynomial is hidden, and your first job is to find it. Every problem has three hints if you get stuck.

Ask yourself first: which symmetric sum does the question really want? Then find a polynomial whose coefficients give it.

Contest cornerreal contest problems

Problem 13.3.1(additional)

AMC 10

The zeros of the function f(x)=x2−ax+2af(x) = x^2 - ax + 2a are integers. What is the sum of the possible values of aa?

Problem 13.3.2(additional)

AIME

Let rr, ss, and tt be the three roots of the equation

8x3+1001x+2008=0.8x^3 + 1001x + 2008 = 0.

Find (r+s)3+(s+t)3+(t+r)3(r + s)^3 + (s + t)^3 + (t + r)^3.

Problem 13.3.3(additional)

AIME

Find the sum of the roots, real and non-real, of the equation x2001+(12−x)2001=0x^{2001} + \left(\frac12 - x\right)^{2001} = 0, given that there are no multiple roots.

Problem 13.3.4(additional)

AHSME

How many distinct ordered triples (x,y,z)(x, y, z) satisfy the following equations?

x+2y+4z=12x + 2y + 4z = 12
xy+4yz+2xz=22xy + 4yz + 2xz = 22
xyz=6xyz = 6

Problem 13.3.5(additional)

AIMEChallenge

Consider the polynomials P(x)=x6−x5−x3−x2−xP(x) = x^6 - x^5 - x^3 - x^2 - x and Q(x)=x4−x3−x2−1Q(x) = x^4 - x^3 - x^2 - 1. Given that z1z_1, z2z_2, z3z_3, and z4z_4 are the roots of Q(x)=0Q(x) = 0, find P(z1)+P(z2)+P(z3)+P(z4)P(z_1) + P(z_2) + P(z_3) + P(z_4).

Problem 13.3.6(additional)

AIMEChallenge

For integers aa, bb, cc and dd, let f(x)=x2+ax+bf(x) = x^2 + ax + b and g(x)=x2+cx+dg(x) = x^2 + cx + d. Find the number of ordered triples (a,b,c)(a, b, c) of integers with absolute values not exceeding 1010 for which there is an integer dd such that g(f(2))=g(f(4))=0g(f(2)) = g(f(4)) = 0.

Problem 13.3.7(additional)

AIMEChallenge

Let P(x)=x2−3x−7P(x) = x^2 - 3x - 7, and let Q(x)Q(x) and R(x)R(x) be two quadratic polynomials also with the coefficient of x2x^2 equal to 11. David computes each of the three sums P+QP + Q, P+RP + R, and Q+RQ + R and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If Q(0)=2Q(0) = 2, then R(0)=mnR(0) = \frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.