Cheatsheet: Polynomials

 Page 1


Polynomials - Class 9 Cheatsheet
1. Introduction to Polynomials
Subtopic Brief
Algebr aic
Expressions
Polynomials are a type of algebr aic expression
involving constants and variables with
oper ations lik e addition, subtr action,
multiplication, and division.
Basic Identities
(x+y)
2
= x
2
+2xy +y
2
, (x-y)
2
= x
2
-2xy +y
2
,
x
2
-y
2
= (x+y)(x-y) . These are used in
factorization and evaluating expressions.
2. Polynomials in One V ariable
Subtopic Brief
Definition
A polynomial p(x) in one variable x is of the form
a
n
x
n
+a
n-1
x
n-1
+···+a
1
x+a
0
, where a
0
,a
1
,...,a
n
are constants, a
n
?= 0 , and exponents are whole
numbers (e.g., x
3
-x
2
+4x+7 ). Expressions lik e
x+
1
x
,
v
x+3 ,
3
v
y +y
2
are not polynomials due to
non-whole number exponents.
T erms and
Coefficients
Each term has a coefficient (e.g., in
-x
3
+4x
2
+7x-2 , coefficients are -1 for x
3
, 4 for
x
2
, 7 for x ,-2 for x
0
). F or x
2
-x+7 , the coefficient
of x is-1 .
Types of
Polynomials
Monomial: One term (e.g., 2x , 5x
5
, u
4
). Binomial:
Two terms (e.g., x+1 , x
2
-x ). Trinomial: Three
terms (e.g., x+x
2
+p ,
v
2+x-x
2
).
Degree of
Polynomial
The highest power of the variable (e.g., x
5
-x
4
+3
has degree 5 , 2-y
2
-y
3
+2y
8
has degree 8 ,
constant 2 has degree 0 ). Zero polynomial (all
coefficients 0 ) has undefined degree.
Polynomial
Classification
Linear: Degree 1 , form ax+b , a?= 0 (e.g., 4x+5 , 2y ,
t+
v
2 ). Quadr atic: Degree 2 , form ax
2
+bx+c ,
a?= 0 (e.g., 2x
2
+5 , 5x
2
+3x+p ). Cubic: Degree 3 ,
form ax
3
+bx
2
+cx+d , a?= 0 (e.g., 4x
3
,
2x
3
+4x
2
+6x+7 ).
1
Page 2


Polynomials - Class 9 Cheatsheet
1. Introduction to Polynomials
Subtopic Brief
Algebr aic
Expressions
Polynomials are a type of algebr aic expression
involving constants and variables with
oper ations lik e addition, subtr action,
multiplication, and division.
Basic Identities
(x+y)
2
= x
2
+2xy +y
2
, (x-y)
2
= x
2
-2xy +y
2
,
x
2
-y
2
= (x+y)(x-y) . These are used in
factorization and evaluating expressions.
2. Polynomials in One V ariable
Subtopic Brief
Definition
A polynomial p(x) in one variable x is of the form
a
n
x
n
+a
n-1
x
n-1
+···+a
1
x+a
0
, where a
0
,a
1
,...,a
n
are constants, a
n
?= 0 , and exponents are whole
numbers (e.g., x
3
-x
2
+4x+7 ). Expressions lik e
x+
1
x
,
v
x+3 ,
3
v
y +y
2
are not polynomials due to
non-whole number exponents.
T erms and
Coefficients
Each term has a coefficient (e.g., in
-x
3
+4x
2
+7x-2 , coefficients are -1 for x
3
, 4 for
x
2
, 7 for x ,-2 for x
0
). F or x
2
-x+7 , the coefficient
of x is-1 .
Types of
Polynomials
Monomial: One term (e.g., 2x , 5x
5
, u
4
). Binomial:
Two terms (e.g., x+1 , x
2
-x ). Trinomial: Three
terms (e.g., x+x
2
+p ,
v
2+x-x
2
).
Degree of
Polynomial
The highest power of the variable (e.g., x
5
-x
4
+3
has degree 5 , 2-y
2
-y
3
+2y
8
has degree 8 ,
constant 2 has degree 0 ). Zero polynomial (all
coefficients 0 ) has undefined degree.
Polynomial
Classification
Linear: Degree 1 , form ax+b , a?= 0 (e.g., 4x+5 , 2y ,
t+
v
2 ). Quadr atic: Degree 2 , form ax
2
+bx+c ,
a?= 0 (e.g., 2x
2
+5 , 5x
2
+3x+p ). Cubic: Degree 3 ,
form ax
3
+bx
2
+cx+d , a?= 0 (e.g., 4x
3
,
2x
3
+4x
2
+6x+7 ).
1
3. Zeroes of a Polynomial
Subtopic Brief
Definition
A real number c is a zero of polynomial p(x) if
p(c) = 0 , also called a root of p(x) = 0 (e.g., for
p(x) = x-1 , p(1) = 0 , so 1 is a zero).
Evaluating
Polynomials
Substitute x with a value (e.g., for
p(x) = 5x
2
-3x+7 , p(1) = 5(1)
2
-3(1)+7 = 9 ; for
q(y) = 3y
3
-4y +
v
11 , q(2) = 16+
v
11 ; for
p(t) = 4t
4
+5t
3
-t
2
+6 , p(a) = 4a
4
+5a
3
-a
2
+6 ).
Finding Zeroes
Solve p(x) = 0 . F or linear polynomial ax+b , zero
is x =-
b
a
(e.g., 2x+1 = 0 , x =-
1
2
; x+5 = 0 ,
x =-5 ). Quadr atic polynomial lik e x
2
-2x has
zeroes 0 and 2 (p(0) = 0 , p(2) = 0 ).
Properties of
Zeroes
Linear polynomials have one zero. Non-zero
constant polynomials have no zeroes. Zero
polynomial has every real number as a zero. A
polynomial can have multiple zeroes (e.g., x
2
-1
has zeroes 1 ,-1 ).
4. F actorisation of Polynomials
Subtopic Brief
F actor Theorem
If p(a) = 0 , then x-a is a factor of p(x) , and vice
versa. F or p(x) = x
3
+3x
2
+5x+6 , p(-2) = 0 , so
x+2 is a facto r . F or 2x+4 , s(-2) = 0 , so x+2 is a
factor .
Finding Constants
If x-a is a factor , solve p(a) = 0 for constant k
(e.g., for p(x) = 4x
3
+3x
2
-4x+k , x-1 is a factor ,
so p(1) = 4+3-4+k = 0 , k =-3 ).
F actorising
Quadr atics
F or ax
2
+bx+c , find numbers p,q such that
p+q = b , pq = ac . F or 6x
2
+17x+5 , p = 2 , q = 15
(2+15 = 17 , 2·15 = 6·5 ), so
6x
2
+2x+15x+5 = (3x+1)(2x+5) . Using F actor
Theorem, test zeroes lik e-
1
3
,-
5
2
. F or y
2
-5y +6 ,
p(2) = 0 , p(3) = 0 , so factors are (y-2)(y-3) .
F actorising Cubics
Find one factor using F actor Theorem, then
divide. F or x
3
-23x
2
+142x-120 , p(1) = 0 , so x-1
is a factor . Divide to get x
2
-22x+120 , factorise as
(x-12)(x-10) , so p(x) = (x-1)(x-10)(x-12) .
2
Page 3


Polynomials - Class 9 Cheatsheet
1. Introduction to Polynomials
Subtopic Brief
Algebr aic
Expressions
Polynomials are a type of algebr aic expression
involving constants and variables with
oper ations lik e addition, subtr action,
multiplication, and division.
Basic Identities
(x+y)
2
= x
2
+2xy +y
2
, (x-y)
2
= x
2
-2xy +y
2
,
x
2
-y
2
= (x+y)(x-y) . These are used in
factorization and evaluating expressions.
2. Polynomials in One V ariable
Subtopic Brief
Definition
A polynomial p(x) in one variable x is of the form
a
n
x
n
+a
n-1
x
n-1
+···+a
1
x+a
0
, where a
0
,a
1
,...,a
n
are constants, a
n
?= 0 , and exponents are whole
numbers (e.g., x
3
-x
2
+4x+7 ). Expressions lik e
x+
1
x
,
v
x+3 ,
3
v
y +y
2
are not polynomials due to
non-whole number exponents.
T erms and
Coefficients
Each term has a coefficient (e.g., in
-x
3
+4x
2
+7x-2 , coefficients are -1 for x
3
, 4 for
x
2
, 7 for x ,-2 for x
0
). F or x
2
-x+7 , the coefficient
of x is-1 .
Types of
Polynomials
Monomial: One term (e.g., 2x , 5x
5
, u
4
). Binomial:
Two terms (e.g., x+1 , x
2
-x ). Trinomial: Three
terms (e.g., x+x
2
+p ,
v
2+x-x
2
).
Degree of
Polynomial
The highest power of the variable (e.g., x
5
-x
4
+3
has degree 5 , 2-y
2
-y
3
+2y
8
has degree 8 ,
constant 2 has degree 0 ). Zero polynomial (all
coefficients 0 ) has undefined degree.
Polynomial
Classification
Linear: Degree 1 , form ax+b , a?= 0 (e.g., 4x+5 , 2y ,
t+
v
2 ). Quadr atic: Degree 2 , form ax
2
+bx+c ,
a?= 0 (e.g., 2x
2
+5 , 5x
2
+3x+p ). Cubic: Degree 3 ,
form ax
3
+bx
2
+cx+d , a?= 0 (e.g., 4x
3
,
2x
3
+4x
2
+6x+7 ).
1
3. Zeroes of a Polynomial
Subtopic Brief
Definition
A real number c is a zero of polynomial p(x) if
p(c) = 0 , also called a root of p(x) = 0 (e.g., for
p(x) = x-1 , p(1) = 0 , so 1 is a zero).
Evaluating
Polynomials
Substitute x with a value (e.g., for
p(x) = 5x
2
-3x+7 , p(1) = 5(1)
2
-3(1)+7 = 9 ; for
q(y) = 3y
3
-4y +
v
11 , q(2) = 16+
v
11 ; for
p(t) = 4t
4
+5t
3
-t
2
+6 , p(a) = 4a
4
+5a
3
-a
2
+6 ).
Finding Zeroes
Solve p(x) = 0 . F or linear polynomial ax+b , zero
is x =-
b
a
(e.g., 2x+1 = 0 , x =-
1
2
; x+5 = 0 ,
x =-5 ). Quadr atic polynomial lik e x
2
-2x has
zeroes 0 and 2 (p(0) = 0 , p(2) = 0 ).
Properties of
Zeroes
Linear polynomials have one zero. Non-zero
constant polynomials have no zeroes. Zero
polynomial has every real number as a zero. A
polynomial can have multiple zeroes (e.g., x
2
-1
has zeroes 1 ,-1 ).
4. F actorisation of Polynomials
Subtopic Brief
F actor Theorem
If p(a) = 0 , then x-a is a factor of p(x) , and vice
versa. F or p(x) = x
3
+3x
2
+5x+6 , p(-2) = 0 , so
x+2 is a facto r . F or 2x+4 , s(-2) = 0 , so x+2 is a
factor .
Finding Constants
If x-a is a factor , solve p(a) = 0 for constant k
(e.g., for p(x) = 4x
3
+3x
2
-4x+k , x-1 is a factor ,
so p(1) = 4+3-4+k = 0 , k =-3 ).
F actorising
Quadr atics
F or ax
2
+bx+c , find numbers p,q such that
p+q = b , pq = ac . F or 6x
2
+17x+5 , p = 2 , q = 15
(2+15 = 17 , 2·15 = 6·5 ), so
6x
2
+2x+15x+5 = (3x+1)(2x+5) . Using F actor
Theorem, test zeroes lik e-
1
3
,-
5
2
. F or y
2
-5y +6 ,
p(2) = 0 , p(3) = 0 , so factors are (y-2)(y-3) .
F actorising Cubics
Find one factor using F actor Theorem, then
divide. F or x
3
-23x
2
+142x-120 , p(1) = 0 , so x-1
is a factor . Divide to get x
2
-22x+120 , factorise as
(x-12)(x-10) , so p(x) = (x-1)(x-10)(x-12) .
2
5. Algebr aic Identities
Subtopic Brief
Basic Identities
(I) (x+y)
2
= x
2
+2xy +y
2
(e.g.,
(x+3)
2
= x
2
+6x+9 ). (II) (x-y)
2
= x
2
-2xy +y
2
(e.g., (4y-1)
2
= 16y
2
-8y +1 ). (III)
x
2
-y
2
= (x+y)(x-y) (e.g.,
25
4
x
2
-
y
2
9
=
(
5
2
x+
y
3
)(
5
2
x-
y
3
)
). (IV)
(x+a)(x+b) = x
2
+(a+b)x+ab (e.g.,
(x-3)(x+5) = x
2
+2x-15 ).
Trinomial Identity
(V)(x+y+z)
2
= x
2
+y
2
+z
2
+2xy+2yz+2zx (e.g.,
(3a+4b+5c)
2
= 9a
2
+16b
2
+25c
2
+24ab+40bc+30ac ,
(4a-2b-3c)
2
= 16a
2
+4b
2
+9c
2
-16ab+12bc-24ac ).
F actorise lik e
4x
2
+y
2
+z
2
-4xy-2yz +4xz = (2x-y +z)
2
.
Cubic Identities
(VI) (x+y)
3
= x
3
+y
3
+3xy(x+y) (e.g.,
(3a+4b)
3
= 27a
3
+64b
3
+108a
2
b+144ab
2
). (VII)
(x-y)
3
= x
3
-y
3
-3xy(x-y) (e.g.,
(5p-3q)
3
= 125p
3
-27q
3
-225p
2
q +135pq
2
).
F actorise lik e8x
3
+27y
3
+36x
2
y+54xy
2
= (2x+3y)
3
.
Sum of Cubes
(VIII)
x
3
+y
3
+z
3
-3xyz = (x+y+z)(x
2
+y
2
+z
2
-xy-yz-zx)
(e.g., 8x
3
+y
3
+27z
3
-18xyz =
(2x+y +3z)(4x
2
+y
2
+9z
2
-2xy-3yz-6xz) ). If
x+y +z = 0 , then x
3
+y
3
+z
3
= 3xyz (e.g.,
(-12)
3
+7
3
+5
3
= 3(-12)(7)(5) ).
Applications
Use identities to evaluate: 105×106 = 11130 ,
(104)
3
= 1124864 , (999)
3
= 997002999 . F actorise
areas/volumes: Area
25a
2
-35a+12 = (5a-3)(5a-4) , V olume
3x
2
-12x = 3x(x-4) .
3