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Mathematical tricks for Problem Solving


This is from the chapter of Theory Of Equations.

(1) The maximum number of positive real roots of a polynomial equation f(x)=0 is the number of changes of signs from positive to negative & negative to positive in f(x).
(2) The number of negative roots of f(x)=0 is the number of changes of signs in f(-x)=0.

USEFUL THEOREM:- Let f(x) be a real polynomial of degree n(≥1) and a,b be two real numbers such that a<b.
(1) If f(a) & f(b) are of positive signs,then polynomial f(x) has at least one and always an odd number of real zeros in (a,b).
(2) If f(a) & f(b) are of the same sign,then polynomial f(x) has either one real zero or an even number of real zeros in (a,b).

Remark:- Every odd degree polynomial has at least one real root since complex roots occur in pairs.

If a>0,prove that x^3 +ax+b =0 has two complex roots.

Sol:- Case-1: b>0
Let f(x)=x^3 +ax+b
Signs are  +,+,+ .
Since there is no sign change in f(x).
Thus,f(x)=0 has no positive root.
Again, f(-x)=-x^3 -ax+b
Signs are -,-,+
Since there is one change of sign in f(-x)
So, f(x)=0 has one -ve real root & two complex roots.
Case-2:- b<0
 Check yourself.
Case-3:- b=0
so,we have ax +x^3 =0
So, x=0, x^2 +a=0 so,it has two complex roots.

Ex.2. Find the number of real roots of the polynomial f(x)= x^5 +x^3 -2x+1.
Sol:- Use Descarte's sign rule
Do yourself.