Polynomials-Types, Graphs, Factor and Remainder theorem

Polynomials-Types, Graphs, Factor and Remainder theorem

Types of polynomials :
i) Constant polynomial : A polynomial having degree  ‘0’ is called constant polynomial
   For ex: \( \sqrt {11} ,3,-7/8,1 \) and so on 
   Degree of  constant  polynomial ‘0’ because \( \frac{{ - 7}} {8} = \frac{{ - 7}} {8} \times 1 = \frac{{ - 7}} {8}x^o \)
ii) Zero polynomial : ‘0’ is called Zero polynomial.  We can’t define degree of zero polynomial 
\( \because \,\,\,\,\,\,0 = 0.x^n + 0.x^{n - 1} + ..... \)
as ‘n’ may be any number.  We can’t say degree 
\(\therefore\)Degree is not defined

iii) Linear polynomial : A polynomial with degree one is called Linear polynomial, Graph of a linear polynomial  is a straight line.

For ex:\( 11x,\,\,\frac{{3x}} {2},\,\,\frac{{\sqrt 2 }} {{x^{ - 1} }} \) \( 3x + 1,\,5p - 3,\,6q - \frac{{13}} {2} \) so on…
iv) Quadratic polynomial :A polynomial of degree ‘2’ is called Quadratic   polynomial 
For ex:\( 3x^2 + 5x + 7,\,x^2 - 5x + 6,\,ax^2 + bx + c(a \ne 0) \) 
Graph of Quadratic polynomial is parabola .

v) Cubic polynomial: A polynomial of degree ‘3’ is called  cubic polynomial
For ex : \( 3x^3 + 7x^2 + x + 1,\,9x^3 + 11x^2 + 20x - 5,\,ax^3 + bx^2 + cx + d(a \ne 0) \) and so on.
v) Biquadratic polynomial : A polynomial of degree ‘4’ is called as a bi-quadratic polynomial.  It is also known as a bi-quadratic polynomial
For ex :\( 5x^4 + 6x^3 + 7x^2 - 11x + 1,\,\sqrt {20} x^4 + 11x^3 - 6x + 1, \)
\( ax^4 + bx^3 + cx^2 + dx + e(a \ne 0) \)                                  
11. Value of a  polynomial : If p(x) is a polynomial at x = K.  The value of polynomial is defined as p(K), it may be zero or any non-zero real number
For ex : 1) For a polynomial \( p(x) = x^2 - 5x + 6 \) at x = 2, \( p(2) = 2^2 - 5.2 + 6 = 0 \)
                                         \( \therefore p(2) = 0 \)
           2) For a polynomial \( p(x) = 8x^3 - 7x^2 + 6x - 5 \) at x = 1
                                      \( p(1) = 8.1^3 - 7.1^2 + 6.1 - 5 \) 
                                     \( = 8 - 7 + 6 - 5 = 2 \)
                                     \( p(1) = 2 \ne 0 \)