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 .*

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 *x*^{2} − 6*x *+ 9 = 0 and *x*^{3} − 7*x*^{2} +159*x *− 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 *x*^{2} − (4 + 2*i*)*x *+ 3 + 4*i *= 0 and *x*^{4} − 8*x*^{3} + 26*x*^{2} − 40*x *+ 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.

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.

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

*ax*^{3} + *bx*^{2} + *cx *+ *d *= 0 .

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

*ax*^{3} + *bx*^{2} + *cx *+ *d *= *a*(*x *− *α** *)(*x *− *β** *)(*x *− *γ** *)

Expanding the right hand side, gives

*ax*^{3} − *a*(*α** *+ *β** *+ *γ** *)*x*^{2} + *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 *a* ≠ 0 and hence division by *a *is meaningful. If a monic cubic
polynomial has roots *α** *, *β** *, and *γ** *, then

coefficient of *x*^{2} = −(*α** *+ *β** *+ *γ** *) ,

coefficient of *x *= *αβ** *+ *βγ** *+ *γα** *, and

constant term = −*αβγ** *.

The same is true for higher degree monic polynomial equations as
well. If a monic polynomial equation of degree *n *has roots *α*_{1} ,*α*_{2} ,...,*α*_{n}* *, then

where Σ*α*_{1} denotes the sum of all roots, Σ*α*_{1}*α*_{2} denotes the sum of
product of all roots taken two at a time, Σ*α*_{1}*α*_{2}*α*_{3} 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 Σ*α*_{1} is written as Σ*α*
, Σ*α*_{1}*α*_{2} is written as Σ*αβ *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 *x*^{n}^{−}^{1} multiplied by (−1) and the product of
the roots is the constant term multiplied by (−1)^{n}* *.

If *α *, *β*, and *γ *are the roots of the
equation *x** ^{3}* +

Since *α *, *β *, and *γ *are the roots of the
equation *x*^{3} + *px*^{2} + *qx *+ *r *= 0 , we have

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 *α*_{1} ,*α*_{2}, …. , *α*_{n}* *? One way of writing a
polynomial equation is multiplication of the factors. That is

( *x *− *α*_{1} )( *x *− *α*_{2} )( *x *− *α*_{3} ) ...( *x *− *αn
*) = 0

is a polynomial equation with roots *α*_{1} ,*α*_{2} ,...,*α** _{n}*. 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

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

*x*^{3} − (*α *+ *β *+ *γ *) *x*^{2} + (*αβ *+ *βγ *+
*γα *) *x *− *αβγ *= 0 .

A polynomial equation of degree *n *with roots *α*_{1} ,*α*_{2} ,K,*α*_{n}* *is given by

where, ∑*α*_{1} , ∑*α*_{1}*α*_{2} , ∑*α*_{1}*α*_{2}*α*_{3} ,K are as defined
earlier.

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

*x*^{3} − (1− 2 + 3) *x*^{2} + (1×(−2) + (−2)× 3 + 3×1) *x *−1×(−2)×
3 = 0

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

Find the sum of the squares of the roots of *ax*^{4} + *bx*^{3} + *cx*^{2} + *dx *+ *e* = 0 , a ≠ 0

Let *α *, *β *,*γ *, and *δ *be
the roots of *ax*^{4} + *bx*^{3} + *cx*^{2} + *dx *+ *e *= 0 .

Then, we get

Find the condition that the roots of cubic equation *x*^{3} + *ax*^{2} + *bx *+ *c*=
0 are in the ratio *p *: *q *: *r *.

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

Then, we get

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

*x*^{3} + *ax*^{2} + *bx *+ *c *= 0 .

Let *α *, *β *, and *γ *be the roots of *x*^{3} + *ax*^{2} + *bx *+ *c* = 0 .

Then, we get

Σ_{1} = α + β + γ = -a , ….(1)

Σ_{2} = αβ + βγ + γα = b , ….(2)

Σ_{3} = αβγ = -c . ….(3)

We have to form the equation whose roots are *α*^{2} , *β*^{2} , and *γ*^{2} .

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

Σ_{1} = *α*^{2} + *β*^{2} + *γ*^{2} = (*α *+ *β *+ *γ *)^{2} - 2(*αβ *+ *βγ
*+ *γα *) = (-*a*)^{2} - 2(*b*) = *a*^{2} - 2*b *,

Σ_{2} = *α*^{2} *β*^{2} + *β*^{2}*γ*^{2} + *γ*^{2}*α*^{2} = (*αβ *+ *βγ *+ *γα *)^{2} - 2((*αβ *)(*βγ
*) + (*βγ *)(*γα *) + (*γα *)(*αβ *))

= (*αβ *+ *βγ *+ *γα *)^{2} - 2*αβγ *(*β *+
*γ *+ *α *) = (*b*)^{2} - 2(-*c*)(-*a*) = *b*^{2} - 2*ca*

Σ_{3} = *α*^{2}*β*^{2}*γ*^{2} = (*αβγ*)^{2} = (-*c*)^{2} = *c*^{2}.

Hence, the required equation is

*x*^{3} – (*α *^{2} + *β*^{2} + *γ*^{2} ) *x*^{2} + (*α *^{2} *β *^{2} + *β*^{2}*γ*^{2} + *γ*^{2}*α*^{2} ) *x *- *α*^{2}*β*^{2}*γ*^{2} = 0.

That is, *x*^{3} - (*a*^{2} - 2*b*) *x*^{2} + (*b*^{2} - 2*ca*) *x
*- *c*^{2} = 0.

If *p *is real, discuss the nature of the roots of the
equation 4*x*^{2} + 4 *px *+ *p *+ 2 = 0 , in terms of *p *

The discriminant Δ=(4 *p*)^{2} - 4 (4)( *p *+
2) = 16 ( *p*^{2} - *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 *<
∞ .

Tags : Definition, Theorem, Formulas, Solved Example Problems | Theory of Equations , 12th Mathematics : UNIT 3 : Theory of Equations

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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

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