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*MATH TRICK-10*

*MATH TRICK-10*

*This is from the chapter of Theory Of Equations.*

**DESCARTE'S RULE OF SIGNS**

(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.

**Ex.1.(FROM RMO PAPER)-**

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

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.