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Polynomial Equation of Degree n

 

Table of Content


Introduction to Polynomial Equation of Degree n

A polynomial equation of degree n is an important topic of the IIT Mathematics syllabus. It comes under the head of Quadratic equations. This is a simple topic which can fetch you direct questions which are too scoring.

Let us first discuss what exactly we mean by a polynomial equation of degree n:

Consider the equation

anxn + an–1xn–1 + an–2xn–2 +……+ a1x + a0 = 0                           … (1)

(a0, a1, … , an are real coefficients and an ≠ 0).

Let a1, a2, ……, an be the roots of equation (1).

Then by simplifying the obtained equations and hence, comparing the coefficients of like powers of x, we get

a1 + a2 + a3 +……+ an = -an-1/ an 

a1a2 + a1a3 + a1a4 + …… + a2a3 +……+ an–1an = an-2/ an,

……………………………………

a1a2 …… ar + … + an–r+1an–r+2 … an = (–1)r an-r/an,

……………………………………

a1a2 …… an = (–1)n a0/an

On the similar lines let us consider an equation ax4 + bx3 + cx2 + dx + e = 0

Let us assume the roots of the equation as a, b, g and d.

Then as derived above, we obtain the following equations:

a + b + g + d = –b/a,

ab + ag + ad + bg + bd + gd = c/a,

abg + abd + agd + bgd = –d/a,

abgd = e/a.

These results are in fact the formulae to be applied while dealing with the numerical based on these.
 

Some Key Points

  • A polynomial equation of degree n has n roots (real or imaginary). 

  • If all the coefficients are real then the imaginary roots occur in pairs i.e. number of complex roots is always even. 

  • If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at least one of the roots will be real.

  • If a is repeated root (repeating r times) of a polynomial equation f(x) = 0 of degree n i.e. f(x) = (x – a)r g(x), where g(x) is a polynomial of degree n – r and g(a) ≠ 0, then f(a) = f’(a) = f’’(a) = … = f(r–1)(a) = 0 and fr(a) ≠ 0.

  • If a polynomial equation of degree n has n + 1 roots say x1, …, xn+1, where xi ≠ xj if i ≠ j, then the polynomial is identically zero i.e. p(x) = 0,  ∀ x ∈ R. this can also be stated as the coefficients a0, ……, an are all zero.

  • If p(a) and p(b) (a < b) are of opposite signs, then p(x) = 0 has odd number of roots in (a, b) i.e. it will have at least one root in (a, b).

  • If coefficients in p(x) have ‘m’ changes in signs, then p(x) = 0 can have at most ‘m’ positive real roots and if p(–x) has ‘t’ changes in sign, then p(x) = 0 can have at most ‘t’ negative real roots. By this we can find maximum number of real roots and minimum number of complex roots of a polynomial equations p(x) = 0.


What is the Remainder Theorem?

The remainder theorem as the name suggests, talks about the remainder when a polynomial is divided by some linear factor. This division obviously results in some quotient as well as remainder. The remainder theorem can be stated as:

When a polynomial say ‘p(x)’ is divided by some linear factor say (x-a), we obtain a remainder of the form p(a). In particular, if p(a) = 0, then (x-a) is a factor of p(x).

The remainder theorem is extremely useful in finding the roots of a polynomial. It helps in identification of a factor which divides the polynomial completely. In large degree polynomials, it would have been a tedious task to find the roots without the remainder theorem. Now, with this theorem we can find out a single root which can then be used synthetically to obtain a smaller polynomial for which the process is repeated until all the roots are obtained.    


You may view the following video for more on remainder theorem:


Let us consider an example based on the remainder theorem:

Illustration: Find all the roots of the equation x4+7x3+3x2-63x-108.

Solution: Seeing the polynomial it is clear that the polynomial will have 5 roots.

We aim to find those values of x for which P(x) = 0.

We first start by substituting x =0, but P (0) = 108 which is not zero and so 0 is not a root of the polynomial. Now we proceed further and try to substitute x = -3. This equals zero and hence, it proves that -3 is a root of the given polynomial. So, by the remainder theorem (x + 3) is a factor of the polynomial. So now we can use synthetic division to get a smaller polynomial

We obtain a new polynomial; let’s call this polynomial f(x)

f(x) = x3+4x2-9x-36.

Now, this is the new reduced polynomial and again we proceed by trial and error. This gives x = 3 as the root and so (x-3) is a factor.

Next we divide f(x) by (x - 3)

We obtain a new polynomial g(x) and a remainder of zero

f(x) = 4x2+7x+12

Repeating the same process again we get two more roots -4 and -3. Hence, the polynomial P(x) can be written as P(x) = (x+3)(x+3)(x-4)(x-3).

Hence, the roots of the polynomial are x = {4, -3, -3, 3}.

Illustration: Let a and b be two roots of the equation x3 + px2 + qx + r = 0 satisfying the relation ab + 1 = 0. Prove that r2 + pr + q + 1 = 0. (r ≠ 0)

Solution: Given equation is x3 + px2 + qx + r = 0.

Let the third root be c, so that abc = –r.                 …… (1)

Given ab = –1

from (1) c = r, which is a root of the given equation.

(r)3 + p(r)2 + q(r) + r = 0 Þ r2 + pr + q + 1 = 0.

Illustration: If b2 < 2ac, then prove that ax3 + bx2 + cx + d = 0 has exactly one real root.

Solution: Let a, b, g be the roots of ax3 + bx2 + cx + d = 0.

Then a + b + g = b/a,

ab + bg + ga = c/a,

abg = -d/a

This gives a2 + b2 + g2 = (a + b + g)2 –2(ab + bg + ga) = b2/a2 –2c/a                                                                            \

= (b2 – 2ac)/ a2

This gives a2 + b2 + g2 < 0, which is not possible if all a, b, g are real. So at least one root is non-real, but complex roots occur in pairs. Hence given cubic equation has two non-real and one real roots.

Alternative Solution:

Let f(x) = ax3 + bx2 + cx + d and f(x) = 0 have all the roots real

So, f’(x) = 3ax2 + 2bx + c = 0 has two real roots.

But its discriminant

= (2b)2 – 4.3.ac = (b2 – 2ac) – ac < 0 (as b2 < 2ac)

which is a contradiction and so f(x) = 0 will not have all the roots real.
 

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