2.11 Sum and Product of Roots (HL)

1. Fundamental Concepts

The Fundamental Theorem of Algebra states that every polynomial of degree $n$ has exactly $n$ complex roots, counted with multiplicity. This means that if $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,\qquad a_n\neq 0,$$ then over $\mathbb{C}$ it can be factorized as $$f(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n),$$ where $r_1,r_2,\ldots,r_n$ are the roots, possibly repeated.

Sum of roots: $$S=r_1+r_2+\cdots+r_n$$
Product of roots: $$P=r_1r_2\cdots r_n$$
The main idea is to compare the coefficients of the expanded factored form with the original polynomial form.

NOTICE

  • The roots may be real or complex.
  • Repeated roots must be counted according to their multiplicity.
  • The formulas below apply even if some of the roots are complex.
  • The leading coefficient must be non-zero: $a_n\neq 0$.

2. Quadratic Functions

Consider the quadratic function $$f(x)=ax^2+bx+c,\qquad a\neq 0.$$ If its roots are $r_1$ and $r_2$, then over $\mathbb{C}$ we may write $$f(x)=a(x-r_1)(x-r_2).$$

Expanding the factored form gives:

$$ \begin{aligned} f(x)&=a(x-r_1)(x-r_2) \\ &=a(x^2-r_1x-r_2x+r_1r_2) \\ &=a\left(x^2-(r_1+r_2)x+r_1r_2\right) \\ &=ax^2-a(r_1+r_2)x+ar_1r_2. \end{aligned} $$

Comparing this with $ax^2+bx+c$, we obtain:

$$ \begin{aligned} b&=-a(r_1+r_2)\quad\Rightarrow\quad r_1+r_2=-\dfrac{b}{a} \\ c&=ar_1r_2\quad\Rightarrow\quad r_1r_2=\dfrac{c}{a} \end{aligned} $$

EXAMPLE 1

Let $$f(x)=2x^2-7x+3.$$ Find the sum and product of the roots.

Solution:

Here $a=2$, $b=-7$, and $c=3$.
The sum of the roots is $$r_1+r_2=-\dfrac{b}{a}=-\dfrac{-7}{2}=\dfrac{7}{2}.$$
The product of the roots is $$r_1r_2=\dfrac{c}{a}=\dfrac{3}{2}.$$
Therefore, $$r_1+r_2=\dfrac{7}{2}\qquad\text{and}\qquad r_1r_2=\dfrac{3}{2}.$$

3. Cubic Functions

Consider the cubic function $$f(x)=ax^3+bx^2+cx+d,\qquad a\neq 0.$$ If its roots are $r_1$, $r_2$, and $r_3$, then $$f(x)=a(x-r_1)(x-r_2)(x-r_3).$$

Expanding the three factors gives:

$$ (x-r_1)(x-r_2)(x-r_3) =x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3. $$
Therefore, $$f(x)=ax^3-a(r_1+r_2+r_3)x^2+a(r_1r_2+r_1r_3+r_2r_3)x-ar_1r_2r_3.$$

Comparing with $ax^3+bx^2+cx+d$, we obtain the cubic Vieta formulas:

$$r_1+r_2+r_3=-\dfrac{b}{a}$$
$$r_1r_2+r_1r_3+r_2r_3=\dfrac{c}{a}$$
$$r_1r_2r_3=-\dfrac{d}{a}$$

EXAMPLE 2

Let $$f(x)=3x^3-5x^2-2x+7.$$ Find the sum of the roots, the sum of all pairwise products, and the product of the roots.

Solution:

Here $a=3$, $b=-5$, $c=-2$, and $d=7$.
Sum of roots: $$r_1+r_2+r_3=-\dfrac{b}{a}=-\dfrac{-5}{3}=\dfrac{5}{3}.$$
Sum of pairwise products: $$r_1r_2+r_1r_3+r_2r_3=\dfrac{c}{a}=-\dfrac{2}{3}.$$
Product of roots: $$r_1r_2r_3=-\dfrac{d}{a}=-\dfrac{7}{3}.$$

4. General Polynomial of Degree $n$

For a general polynomial $$f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+a_0,\qquad a_n\neq 0,$$ with roots $r_1,r_2,\ldots,r_n$, we write $$f(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n).$$

Matching coefficients gives the following general relations:

Sum of all roots: $$a_{n-1}=-a_n(r_1+r_2+\cdots+r_n)$$ and hence $$r_1+r_2+\cdots+r_n=-\dfrac{a_{n-1}}{a_n}.$$
Product of all roots: $$a_0=(-1)^n a_n(r_1r_2\cdots r_n)$$ and hence $$r_1r_2\cdots r_n=(-1)^n\dfrac{a_0}{a_n}.$$
Intermediate coefficient relation: the sum of all pairwise products of roots is $$\sum_{1\leq i < j\leq n}r_i r_j=\dfrac{a_{n-2}}{a_n}.$$
More generally, the elementary symmetric sums alternate signs: $$f(x)=a_n\left[x^n-S_1x^{n-1}+S_2x^{n-2}-S_3x^{n-3}+\cdots+(-1)^nS_n\right],$$ where $$ \begin{aligned} S_1&=\sum_i r_i, \\ S_2&=\sum_{i < j}r_ir_j, \\ S_3&=\sum_{i < j < k}r_ir_jr_k, \\ \ &\vdots \\ S_n&=r_1r_2\cdots r_n. \end{aligned} $$

MONIC POLYNOMIALS

If the leading coefficient is $1$, then the polynomial is called monic. For example, $$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0.$$ In this case the formulas simplify:

$$r_1+r_2+\cdots+r_n=-a_{n-1}$$
$$r_1r_2\cdots r_n=(-1)^n a_0$$

5. Analytical Applications

EXAMPLE 3 (Cubic Application)

Let $$f(x)=2x^3+ax^2+bx+c.$$ It is given that the sum of the roots is $3.5$, the product of the roots is $-5$, and the polynomial is divisible by $(x+2)$. Find $a$, $b$, and $c$.

Solution:

Since this is a cubic polynomial with leading coefficient $2$, the sum formula gives $$r_1+r_2+r_3=-\dfrac{a}{2}.$$ The given sum is $3.5=\dfrac{7}{2}$, so $$-\dfrac{a}{2}=\dfrac{7}{2}\quad\Rightarrow\quad a=-7.$$
The product formula for a cubic is $$r_1r_2r_3=-\dfrac{c}{2}.$$ The given product is $-5$, so $$-\dfrac{c}{2}=-5\quad\Rightarrow\quad c=10.$$
Because $f(x)$ is divisible by $(x+2)$, the Factor Theorem gives $$f(-2)=0.$$
Substitute $a=-7$ and $c=10$ into the polynomial: $$f(x)=2x^3-7x^2+bx+10.$$ Then $$f(-2)=2(-2)^3-7(-2)^2+b(-2)+10.$$
Hence $$-16-28-2b+10=0.$$ Therefore, $$-34-2b=0\quad\Rightarrow\quad b=-17.$$
Final answer: $$a=-7,\qquad b=-17,\qquad c=10.$$

EXAMPLE 4 (Quartic Application)

Let $$f(x)=ax^4-10x^3+bx+c.$$ The sum of the roots is $2$, the product of the roots is $5$, and the polynomial is divisible by $(x-1)$. Find $a$, $b$, and $c$.

Solution:

This is a quartic polynomial, so $n=4$. The leading coefficient is $a$, and the coefficient of $x^3$ is $-10$.
Using the sum formula, $$r_1+r_2+r_3+r_4=-\dfrac{a_3}{a_4}.$$ Here $a_3=-10$ and $a_4=a$, so $$2=-\dfrac{-10}{a}=\dfrac{10}{a}.$$ Therefore, $$a=5.$$
Since $n=4$, the product formula gives $$r_1r_2r_3r_4=(-1)^4\dfrac{a_0}{a_4}=\dfrac{c}{a}.$$ The given product is $5$, so $$5=\dfrac{c}{5}\quad\Rightarrow\quad c=25.$$
Because $(x-1)$ is a factor, the Factor Theorem gives $$f(1)=0.$$
Substitute $a=5$ and $c=25$: $$f(x)=5x^4-10x^3+bx+25.$$ Then $$f(1)=5-10+b+25=0.$$
Therefore, $$20+b=0\quad\Rightarrow\quad b=-20.$$
Final answer: $$a=5,\qquad b=-20,\qquad c=25.$$

EXAMPLE 5 (Constructing a Polynomial from Roots)

A monic cubic polynomial has roots $2$, $-1$, and $4$. Find the polynomial.

Solution:

Since the roots are $2$, $-1$, and $4$, the factored form is $$f(x)=(x-2)(x+1)(x-4).$$
First multiply two factors: $$(x-2)(x+1)=x^2-x-2.$$
Now multiply by the third factor: $$f(x)=(x^2-x-2)(x-4).$$
Expanding: $$f(x)=x^3-4x^2-x^2+4x-2x+8.$$
Therefore, $$f(x)=x^3-5x^2+2x+8.$$
Check using Vieta: $$2+(-1)+4=5,$$ so the coefficient of $x^2$ should be $-5$, which agrees with the result.