class 9 Polynomials.

2. Introduction

3. Degree of a Polynomial in one variable.
5 3 2 x 

4. Degree of a Polynomial in two variables.
• What is degree of the following polynomial?
5 7 3 9 4 2 3 3 x y  x  xy  y 
• The answer is five because if we add 2 and 3 , the answer is
five which is the highest power in the whole polynomial.
3 5 8 2 9 3 4 2 x y  x  xy  y 
E.g.- is a polynomial
in x and y of degree 7.

5. Polynomials in one variable

6. Polynomials in one variable
The degree of a polynomial in one variable is the
largest exponent of that variable.
A constant has no variable. It is a 0 degree
polynomial.
2
4x 1
This is a 1st degree polynomial. 1st degree
polynomials are linear.
5 2 14 2 x  x  This is a 2nd degree polynomial.
2nd degree polynomials are
quadratic.
3 18 3 x 
This is a 3rd degree polynomial. 3rd
degree polynomials are cubic.

7. Examples
Polynomials Degree Classify by degree Classify by no. of
Text
Text
Text
terms.
5 0 Constant Monomial
2x - 4 1 Linear Binomial
3x2 + x 2 Quadratic Binomial
x3 - 4x2 + 1 3 Cubic Trinomial

8. Standard Form
Phase 1 Phase 2
To rewrite a
polynomial in
standard form,
rearrange the terms
of the polynomial
starting with the
largest degree term
and ending with the
lowest degree term.
The leading coefficient, the
coefficient of the first term
in a polynomial written in
standard form, should be
positive.

9. Remainder Theorem

10. Questions on Remainder Theorem
Q.) Find the remainder when the polynomial
f(x) = x4 + 2x3 – 3x2 + x – 1 is divided by (x-2).
A.) x-2 = 0 x=2
By remainder theorem, we know that when f(x) is divided by (x-2), the
remainder is x(2).
Now, f(2) = (24 + 2*23 – 3*22 + 2-1)
= (16 + 16 – 12 + 2 – 1) = 21.
Hence, the required remainder is 21.

11. Factor Theorem

12. Algebraic Identities
Some common identities used to factorize polynomials
(a+b)2=a2+b2+2ab (a-b) (x+a)(x+b)= x 2 + (a+b)x + ab 2=a2+b2-2ab a2-b2=(a+b)(a-b)

13. Algebraic Identities
Advanced identities used to factorize polynomials
(x-y)3=x3-y3-
3xy(x-y)
(x+y+z)2=x2+y2+z2+2
xy+2yz+2zx
x3 + y3 + z3 –
3xyz = (x + y +
z)(x2 + y2 + z2 –
xy – yz – zx)
(x+y)3=x3+y3+3x
y(x+y)
x3+y3=(x+y) *
(x2+y2-xy)
x3-y3=(x+y) *
(x2+y2+xy)
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