THEORY OF POLYNOMIAL EQUATIONS
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.
anxn + an – 1xn – 1 + an – 2xn – 2 + ….. + a1x + a0 º an(x - a1) (x - a2) ….. (x - an) ….(2).
Comparing the coefficients of like powers of x, we get
a1 + a2 + a3 + …. + an = -
a1a2 + a1a3 + a1a4 + …. + a2a3 + … + an - 1an = 
………………………………
a1a2 . . . . .ar + …. + an-r+1an-r+2 … an = ( -1)r
…….………………………….
a1a2 … an = (-1)n
In general åa1.a2......ai = (–1)i
e.g. If a, b, g and d are the roots of ax4 + bx3 + cx2 + dx + e = 0 then.
a + b + g + d = -b/a
ab + ag+ ad + bg + bd + gd = c/a
abg + abd + agd + bgd = -d/a
abgd = e/a
Remarks:
- 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 f r (a) > 0 and vice versa.
- A polynomial in x can be factorized into real factors in the following way:
- Determine all roots x1,...,xn of the polynomial in x and write in form as in (2). If some
roots are complex conjugates (say x1 º c + id, x2 º c - id), then combine them as
(x - x1)(x - x2) = (x - c)2 + d2. - Thus polynomial in x of degree n can be factorized into a product of linear/quadratic form.
Remainder Theorem
- If we divide polynomial in x by (x - a) , the remainder obtained is p(a). Note that If p(a) = 0, then x - a is a factor of p(x).
- If a polynomial of degree n has n + 1 roots say x1,...xn + 1 , xi > xj if i > j, then the polynomial is identically zero. ie. p(x) = O "x ∈ R.
(In other words, the coefficients a0,...an are all zero).
Example 1
If the sum of the roots of the equation ax2 + bx + c = 0 is equal to sum of the squares of their reciprocals, then show that ab2, a2c, bc2
are in A.P.
Solution:
Let a, b be the roots of the given equation. Then .
&⇒ 2a2c = b2a + c2 b
&⇒ ab2, a2c, bc2 are in A.P.
Example 2
If a is a root of the equation ax2+bx+c=0 and b is a root of the equation -ax2 + bx + c = 0, then prove that there will be a root of the equation 
lying between a and b.
Solution:
Let f(x) = 
f(a) = 
a2 +ba +c = aa2 +ba +c - a2
= - a2 (As a is a root of ax2 +bx +c = 0 )
And f(b) = b2 +bb+c = -ab2 + bb +c +
ab2
= ab2 ( As b is a root of - ax2 +bx +c = 0
Now f(a). f(b) = 
a2a2b2 < 0
&⇒ f(x) = 0 has one real root between a and b.
Example 3
Let P(x)º
Prove that P(x) has the property that P(y) = y2 for all y ∈ R. .
Solution:
Note that P(a) = a2, P(b) = b2 and P(c) = c2.
Consider the polynomial Q(x) = P(x) - x2 , Q(x) has degree atmost 2.
Also Q(a) = Q(b) = Q(c) = 0
&⇒ Q(x) has 3 distinct roots.
It follows that Q(x) is identically zero
ie. Q(y) = 0 "y ∈ R.
&⇒ P(y) -y2 = 0 "y∈ R
&⇒ P(y) = y2 "y∈ R
· If p(a) and p(b) (a < b) are of opposite sign, then p(x) = 0 have odd number of roots in (a, b), i.e. it have at least one root in (a, b).
· If coefficients in p(x) have `m’ changes in signs, then p(x) = 0 have at most `m’ +ve real roots and if p(-x) have `t’ changes in sign, then p(x) = 0 have at most `t’ negative real roots. By this we can find the maximum number of real roots and also the minimum number of complex roots of a polynomial equations p(x) = 0.
Example 4
Find the number of maximum possible positive / negative roots of the equation x6 - 3x5 + 4x3 + 3x2 + 4 = 0.
Solution:
Let f(x) = x6 - 3x5 +4x3 + 3x2 + 4
Here f(x) has 2 changes in sign, first change is from positive to negative for x6 to x5 and second change is from negative to positive for x5 to x3 . So f(x) = 0 has atmost 2 positive real roots. .
f(-x) = x6 + 3x5 - 4x3 + 3x2+4 has 2 changes in sign so f(x) = 0 has atmost 2 negative real roots. Since the least degree in f(x) is zero, so f(x) has no root as x = 0 . Hence f(x) = 0 has atmost four real roots, therefore atleast two complex roots.
Example 5
If b2 < 2ac, then prove that ax3 + bx2 + cx + d = 0 has exactly one real roots.
Solution:
Let a, b, g be the roots of ax3 +bx2 +cx +d =0
Then a + b + g = 
ab + bg + ga = 
abg = 
a2 +b2 +g2 = (a + b + g )2 – 2(ab + bg +ga)
= 
&⇒
