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# Vieta’s formula for Polynomial Equations

What we have learnt for quadratic polynomial, can be extended to polynomials of higher degree.

Vieta’s formula for Polynomial Equations

What we have learnt for quadratic polynomial, can be extended to polynomials of higher degree. In this section we study the relations of the zeros of a polynomial of higher degree with its coefficients. We also learn how to form polynomials of higher degree when some information about the zeros are known. In this chapter, we use either zeros of a polynomial of degree n or roots of polynomial equation of degree n .

## (a) The Fundamental Theorem of Algebra

If  a  is  a  root  of  a  polynomial  equation P(x) = 0 ,  then  (x a)  is  a  factor of P(x) . So, deg (P(x)) 1. If a and b are roots of P(x) = 0 then (x a)(x b) is a factor of P(x) and hence deg (P(x)) 2 . Similarly if P(x) = 0 has n roots, then its degree must be greater than or equal to n. In other words, a polynomial equation of degree n cannot have more than n roots.

In earlier classes we have learnt about “multiplicity”. Let us recall what we mean by “multiplicity”.

We know if (x a)k  is a factor of a polynomial equation P(x) = 0 and (x a)k +1 is not a factor of the polynomial equation, P(x) = 0 , then a is called a root of multiplicity k . For instance, 3 is a root of multiplicity 2 for the equation x2 6x + 9 = 0 and x3 7x2 +159x 9 = 0 . Though we are not going to use complex numbers as coefficients, it is worthwhile to mention that the imaginary number 2 + i is a root of multiplicity 2 for the polynomials x2 (4 + 2i)x + 3 + 4i = 0 and x4 8x3 + 26x2 40x + 25 = 0.

If a is a root of multiplicity 1 for a polynomial equation, then a is called a simple root of the polynomial equation.

If P ( x) = 0 has n roots counted with multiplicity, then also, we see that its degree must be greater than or equal to n . In other words, “a polynomial equation of degree n cannot have more than n roots, even if the roots are counted with their multiplicities”.

One of the important theorems in the theory of equations is the fundamental theorem of algebra.

As the proof is beyond the scope of the Course, we state it without proof.

## Theorem 3.1 (The Fundamental Theorem of Algebra)

Every polynomial equation of degree n ≥ 1 has at least one root in C.

Using this, we can prove that a polynomial equation of degree n has at least n roots in C when the   roots are counted with their multiplicities. This statement together with our discussion above says that

a polynomial equation of degree n has exactly n roots in C when the roots are counted with their multiplicities.

Some authors state this statement as the fundamental theorem of algebra.

## (b) Vieta’s Formula

### (i) Vieta’s Formula for Polynomial equation of degree 3

Now we obtain these types of relations to higher degree polynomials. Let us consider a general cubic equation

ax3 + bx2 + cx + d = 0 .

By the fundamental theorem of algebra, it has three roots. Let α , β , and γ be the roots. Thus we have

ax3 + bx2 + cx + d = a(x α )(x β )(x γ )

Expanding the right hand side, gives

ax3 a(α + β + γ )x2 + a(αβ + βγ + γα )x a(αβγ ) .

Comparing the coefficients of like powers, we obtain

α + β + γ = −b/a , αβ + βγ + γα = c/a and αβγ = −d/a

Since the degree of the polynomial equation is 3, we have a0 and hence division by a is meaningful. If a monic cubic polynomial has roots α , β , and γ , then

coefficient of x2 = −(α + β + γ ) ,

coefficient of x = αβ + βγ + γα , and

constant term = αβγ .

### (ii) Vieta’s Formula for Polynomial equation of degree n > 3

The same is true for higher degree monic polynomial equations as well. If a monic polynomial equation of degree n has roots α1 ,α2 ,...,αn , then where Σα1 denotes the sum of all roots, Σα1α2 denotes the sum of product of all roots taken two at a time,  Σα1α2α3 denotes the sum of product of all roots taken three at a time, and so on. If  α , β ,γ , and δ are the roots of a quartic equation, then Σα1 is written as Σα ,  Σα1α2   is written as  Σαβ and so on. Thus we have,

Σα = α + β + γ + δ

Σαβ = αβ + αγ + αδ + βγ + βδ + γδ

Σαβγ = αβγ + αβδ + αγδ + βγδ

Σαβγδ = αβγδ

When the roots are available in explicit numeric form, then also we use these convenient notations. We have to be careful when handling roots of higher multiplicity. For instance, if the roots of a cubic equation are 1, 2, 2, then ∑α = 5 and ∑αβ = (1× 2) + (1× 2) + (2 × 2) = 8 .

From the above discussion, we note that for a monic polynomial equation, the sum of the roots is the coefficient of xn1 multiplied by (−1) and the product of the roots is the constant term multiplied by  (−1)n  .

### Example 3.3

If α , β, and γ are the roots of the equation x3 + px2 + qx + r = 0 , find the value of in terms of the coefficients.

### Solution

Since α , β , and γ are the roots of the equation x3 + px2 + qx + r = 0 , we have ## (c) Formation of Polynomial Equations with given Roots

We have constructed quadratic equations when the roots are known. Now we learn how to form polynomial equations of higher degree when roots are known. How do we find a polynomial equation of degree n with roots α1 ,α2, …. , αn ? One way of writing a polynomial equation is multiplication of the factors. That is

( x α1 )( x α2 )( x α3 ) ...( x αn ) = 0

is a polynomial equation with roots α1 ,α2 ,...,αn. But it is not the usual way of writing a polynomial equation. We have to write the polynomial equation in the standard form which involves more computations. But by using the relations between roots and coefficients, we can write the polynomial equation directly; moreover, it is possible to write the coefficient of any particular power of x without finding the entire polynomial equation.

A cubic polynomial equation whose roots are α , β , and γ is

x3 − (α + β + γ ) x2 + (αβ + βγ + γα ) x αβγ = 0 .

A polynomial equation of degree n with roots α1 ,α2 ,K,αn is given by where,  ∑α1 , ∑α1α2 , ∑α1α2α3 ,K  are as defined earlier.

For instance, a polynomial equation with roots 1, −2 , and 3 is given by

x3 − (1− 2 + 3) x2 + (1×(−2) + (−2)× 3 + 3×1) x −1×(−2)× 3 = 0

which, on simplification, becomes x3 − 2x2 − 5x + 6 = 0 . It is interesting to verify that the expansion  of ( x −1)( x + 2)( x − 3) = 0 is x3 − 2x2 − 5x + 6 = 0 .

### Example 3.4

Find the sum of the squares of the roots of ax4 + bx3 + cx2 + dx + e  = 0 , a ≠ 0

### Solution

Let   α , β ,γ ,  and  δ be the roots of ax4 + bx3 + cx2 + dx + e = 0 .

Then, we get ### Example 3.5

Find the condition that the roots of cubic equation x3 + ax2 + bx + c= 0 are in the ratio p : q : r .

### Solution

Since roots are in the ratio p : q : r , we can assume the roots as pλ, qλ and rλ,

Then, we get ### Example 3.6

Form the equation whose roots are the squares of the roots of the cubic equation

x3 + ax2 + bx + c = 0 .

### Solution

Let α , β , and γ be the roots of x3 + ax2 + bx + c = 0 .

Then, we get

Σ1   = α + β + γ = -a , ….(1)

Σ2  = αβ + βγ + γα = b , ….(2)

Σ3   = αβγ = -c . ….(3)

We have to form the equation whose roots are α2 , β2 , and γ2 .

Using (1), (2) and (3), we find the following:

Σ1 = α2 + β2 + γ2 = (α + β + γ )2 - 2(αβ + βγ + γα ) = (-a)2 - 2(b) = a2 - 2b ,

Σ2 = α2 β2  + β2γ2 + γ2α2 = (αβ + βγ + γα )2 - 2((αβ )(βγ ) + (βγ )(γα ) + (γα )(αβ ))

= (αβ + βγ + γα )2 - 2αβγ (β + γ + α ) = (b)2 - 2(-c)(-a) = b2 - 2ca

Σ3 = α2β2γ2 = (αβγ)2 = (-c)2 = c2.

Hence, the required equation is

x3 – (α 2 + β2 + γ2 ) x2 + (α 2 β 2 + β2γ2 + γ2α2 ) x - α2β2γ2 = 0.

That is,   x3  - (a2  - 2b) x2  + (b2  - 2ca) x - c2  = 0.

### Example 3.7

If p is real, discuss the nature of the roots of the equation 4x2 + 4 px + p + 2 = 0 , in terms of

### Solution

The discriminant  Δ=(4 p)2  - 4 (4)( p + 2) = 16 ( p2  - p - 2) = 16 ( p +1)( p - 2) . So, we get

Δ < 0 if -1 < p < 2

Δ = 0 if p = -1 or p = 2

Δ > 0 if -∞ < p < -1 or 2 < p < ∞

Thus the given polynomial has

imaginary roots if -1 < p < 2 ;

equal real roots if p = -1 or p = 2 ;

distinct real roots if -∞ < p < -1 or 2 < p < ∞ .

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12th Mathematics : UNIT 3 : Theory of Equations : Vieta’s formula for Polynomial Equations | Definition, Theorem, Formulas, Solved Example Problems | Theory of Equations