# Polynomial Curve

Introduction :

Let f(x) be a polynomial in ‘x’ of degree = 1whose co - efficient are real or complex number. Then f(x)=0 is said to be an algebraic equation or a polynomial curve. The degree of the polynomial f(x)=0.

A curve of degree 1 is f(x) =a0x+a1 where a0?0  .

A curve of degree 2 is f(x)= a0x2+a1x+a2 where a0?0,

A curve of degree 3 is

f(x)= a0x3+a1x^2+a3

where a0?0, this called a cubic curve.

The general form of an algebraic equation of degree ‘n’ is

f=(x)= b0xn+b1xn-1+…………….+bn-1+bn

whewhere b0 , b1, ……………………….. bn are real or complex numbers and b0?0
Properties of Polynomial Curve :

If ‘a’ be a value of x for which f(x) become zero, i.e. if f(a)=0 then ‘a’ is said to be a root of the curve f(x) =0

Polynomial curve

2)      Fundamental theorem: Every polynomial curve has a root, real or complex.

3)      A curve of degree ‘n’ has exactly ‘n’ root.

4)      If a curve with real coefficients has a complex root (a+ib) then it has also the conjugate complex root (a-ib).
Problem for Polynomial Curve:

Ex 1):Solve the cure x^4 + x^2 +2x+6 =0, it is given that (l+i) is root.

Sol: Let f(x) = x^4 + x^4-2x+6.

Step 1:Since f(x) =0 is a curve with real coefficient and (1+i) is a root of equation, (1+i) is also a root.

Therefore

Step 2:={x-(1+i)} {x-(1-i)}

=(x-1-I0(x-1+i)

=(x-1)^2-i2       [since (a+b) (a-b) =a2-b2]

Step 3: Simplify

=x^2-2x+1+1     [since i2=-1]

=x^2-2x+2 is a factor of f(x).

Step 4:Now,

x^4 + x^2 -2x +6

= x^2 (x^2 -2x+2) +2x(x^2 -2x+2) +3(x^2 +2x +3)

Step 5: Solve for x . `rArr` x^2 + 2x +3 =0

Gives, x= -1 ±v2 i

Therefore the roots are 1± I , -1?v2i.

Ex 2): Show that 2 is a root of quadratic curve. F(x)=   x^2 - 3x +2

Sol:Step 1: The given quadratic curve is

X2 - 3x +2

Step 2:If ‘2’ satisfy the given quadratic curve then ‘2’ is a root of given expression.

Now, putting x=2 we get ,

22 – 3(2) +2

=4 -6 +2

=6-6

=0.