Consider the equation

a_{n}x^{n} + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} + …. + a_{1}x + a_{0} = 0 . . . . (1).

(a_{0}, a_{1}…., a_{n} are real coefficients and a_{n} > 0).

Let a_{1}, a_{2},….,a_{n} be the roots of equation (1). Then.

a_{n}x^{n} + a_{n – 1}x^{n – 1} + a_{n – 2}x^{n – 2} + ….. + a_{1}x + a_{0} º a_{n}(x - a_{1}) (x - a_{2}) ….. (x - a_{n}) ….(2).

Comparing the coefficients of like powers of x, we get

a_{1} + a_{2} + a_{3} + …. + a_{n} = -

a_{1}a_{2} + a_{1}a_{3} + a_{1}a_{4} + …. + a_{2}a_{3} + … + a_{n - 1}a_{n} =

………………………………

a_{1}a_{2} . . . . .a_{r} + …. + a_{n-r+1}a_{n-r+2} … a_{n }= ( -1)^{r}

…….………………………….

a_{1}a_{2} … a_{n} = (-1)^{n}

In general åa_{1}.a_{2}......a_{i} = (–1)^{i}

e.g. If a, b, g and d are the roots of ax^{4} + bx^{3} + cx^{2} + 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 x
_{1},...,x_{n}of the polynomial in x and write in form as in (2). If some

roots are complex conjugates (say x_{1}º c + id, x_{2}º c - id), then combine them as

(x - x_{1})(x - x_{2}) = (x - c)^{2}+ d^{2}. - Thus polynomial in x of degree n can be factorized into a product of linear/quadratic form.

- 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 x
_{1},...x_{n + 1}, x_{i}> x_{j}if i > j, then the polynomial is identically zero. ie. p(x) = O "x ∈ R.

(In other words, the coefficients a_{0},...a_{n} are all zero).

If the sum of the roots of the equation ax^{2} + bx + c = 0 is equal to sum of the squares of their reciprocals, then show that ab^{2}, a^{2}c, bc^{2}

are in A.P.

Let a, b be the roots of the given equation. Then .

&⇒ 2a^{2}c = b^{2}a + c^{2} b

&⇒ ab^{2}, a^{2}c, bc^{2} are in A.P.

If a is a root of the equation ax^{2}+bx+c=0 and b is a root of the equation -ax^{2 }+ bx + c = 0, then prove that there will be a root of the equation lying between a and b.

Let f(x) =

f(a) = a^{2} +ba +c = aa^{2} +ba +c - a^{2}

= - a^{2 }(As a is a root of ax^{2} +bx +c = 0 )

And f(b) = b^{2} +bb+c = -ab^{2} + bb +c +ab^{2}

= ab^{2} ( As b is a root of - ax^{2} +bx +c = 0

Now f(a). f(b) = a^{2}a^{2}b^{2} < 0

&⇒ f(x) = 0 has one real root between a and b.

Let P(x)º

Prove that P(x) has the property that P(y) = y^{2} for all y ∈ R. .

Note that P(a) = a^{2}, P(b) = b^{2} and P(c) = c^{2}.

Consider the polynomial Q(x) = P(x) - x^{2} , 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) -y^{2} = 0 "y∈ R

&⇒ P(y) = y^{2} "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.

Find the number of maximum possible positive / negative roots of the equation x^{6} - 3x^{5} + 4x^{3} + 3x^{2} + 4 = 0.

Let f(x) = x^{6} - 3x^{5} +4x^{3} + 3x^{2} + 4

Here f(x) has 2 changes in sign, first change is from positive to negative for x^{6} to x^{5} and second change is from negative to positive for x^{5} to x^{3} . So f(x) = 0 has atmost 2 positive real roots. .

f(-x) = x^{6} + 3x^{5} - 4x^{3} + 3x^{2}+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.

If b^{2} < 2ac, then prove that ax^{3} + bx^{2} + cx + d = 0 has exactly one real roots.

Let a, b, g be the roots of ax^{3} +bx^{2} +cx +d =0

Then a + b + g =

ab + bg + ga =

abg =

a^{2} +b^{2} +g^{2} = (a + b + g )^{2} – 2(ab + bg +ga)

=

&⇒