POLYNOMIAL
Polynomial : Polynomial means an algebriac sum made up of many terms involving powers of the variable. A polynomial of degree n is defined as an expression
of the a xn + a xn-1 + a xn-2 +…… a
x+a
where a ,a ,a
….. an are real numbers or
0 1 2
a-1 n
0 1 2
coefficients of the polynomial, n is a integer and a0 ¹ 0.
Example : (1) 3x – 1 is a polynomial of degree one.
- Ö5 x2 – 3x + 2 is a polynomial of degree
- 5x3 – 2x2 + x – 6 is a polynomials of degree
Example 1 and 3 are polynomials with integral coefficients and (2) is a polynomal with non-integral coefficients.
Zero polynomial : These are the polynomials in which all coefficients are zero.
Monomial : A polynomial that has only one term e.g. 4×2, 3×2, 2x, 5 etc.
Bionomial : A polynomial which has two terms e.g. 5x2 + 2, 3x2 + 1 etc.
Divisor : A term d(x) is called a divisor of a polynomiall p(x), if d(x) is a factor of p(x).
In other words, a factor d(x) is a divisor of a polynomiall p(x) if there exists a polynomial q(x) such that p(x) = d(x) q(x).
Example : Let p(x) = x2 – 5x + 6. Then, p(x) = (x – 2) (x – 3). Clearly, (x – 2) and (x
-2) (x – 3) are factors of p(x). Therefore (x-2) and (x – 3) ar divisors of p(x).
Greatest common divisor : Consider two polynomials p(x) = (x – 2)2 (x – 3) (x + 3)2 and q(x) = (x – 2)3 (x – 3)2 (x + 5). We observe that among all the common divisors, (x – 2)2 (x – 3) are two common divisors of highest degree. The divisor – (x – 2)2 (x – 3) has its highest degree term coefficient as negative whereas the divisor (x – 2)2 (x – 3) has its highest degree term coeffiient positive. Such a divisor is called the highest common divisor or greatest common divisor as defined below.
Greatest Common Divisor (GCD) or Highest Common Factor (HCF) : The greatest common divisor (g.c.d) of two polynomials p(x) and q(x) is that common divisor which has highest degree among all common divisors and in which the coefficient of highest degree term is positive.
The g.c.d. of two polynomials can be obtained by using the following algorithm.
ALGORITHM FOR FINDING G.C.D. BY FACTORISATION METHOD
Step 1. Obtain two polynomials. Let the polynomials be p(x) and q(x).
Step 2. Factorise the polynomials p(x) and q(x).
Step 3. Express p(x) and q(x) in the form of product of powers of different factors.
Step 4. Determine the maximum power of each factor which is common to the given polynomials.
Step 5. Get the product of all factors obtained in step 4.
The product obtained in step 5 is the g.c.d. of the given polynomials.
The following formulae are useful for factorizing the polynomials (A + B) (A – B) = A2 – B2
(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA (A + B)0 = 1
(A – B)0 = 1
(A + B)2 = A2 + B2 + 2AB (A – B)2 = A2 + B2 – 2AB
(A + B)3 = A3 + B3 + 3A2B + 3AB2
(A – B)3 = A3 + B3 – 3A2B + 3AB2 A3 + B3 = (A + B) (A2 – AB + B2) A3 – B3 = (A – B) (A2 + B2 + AB)
Example : Find the g.c.d. (x + 2)2 (x – 3)2 (x + 3)2 and (x + 3)3 (x + 2)3 (x – 3).
Solution : Let p(x) = (x + 2)2 (x – 3)2 (x + 3)2 and q(x) = (x + 3)3 (x + 2)3 (x + 3). The highest degree common divisors of two polynomials are (x + 2)2, (x – 3) and (x + 3)2. Therefore, g.c.d. of p(x) and q(x) is (x + 2)2 (x – 3) (x + 3)2.
If p(x) and q(x) are two polynomials then we define
l.c.m. of p(x) and q(x)
= p(x).q(x)
g.c.d. of p(x) and q(x)
It follows from this definition that the l.c.m. of two polynomials p(x) and q(x) is the polynomial of lowest degree which has p(x) and q(x) as divisors and whose coefficient of highest degree term has same sign as the sign of the coefficient of highest. degree term in the product p(x).q(x).
Example : Find the l.c.m. 2x2 + 5x – 3 and 2x2 – 7x + 3.
Solution : Let p(x) = 2×2 + 5x-3 and q(x) = 3×2 – 7x + 2.
Then, p(x) = 2x2 + 5x – 3 = 2x2 + 6x – x – 3 = 2x (x + 3) – (x + 3) = (2x – 1) (x + 3) and q(x) = 2x2 – 7x + 3 = 2x2 – 6x – x + 3 = 2x (x – 3) – (x – 3) = (2x – 1) (x – 3)
So, g.c.d. of p(x) and q(x) is (2x-1). Hence, l.c.m. of p(x) and q(x)
= p(x).q(x)
g.c.d. of p(x) and q(x)
= (2x – 1) (x + 3). (2x – 1) (x -3)
2x – 1
= (x + 3) (x -3).
- HCF or C.D. : HCF of two or more polynomials is the common factor of highest degree.
- C.M. : LCM of two or more polynomials is the common multiple of least degree.
(3) L.C.M.
Product of two polynomial HCF of two polynomials
REMAINDER AND FACTOR THEOREMS
Remainder theorem : Let f(x) be a polynomial of degree greater than or equal to one and a be any real number. If f(x) is divisible by (x – a), then the remainder is equal to f(a).
Example : Determine the remainder when the polynomial f(x) = x3 – 3x2 + 2x + 1 is divided by (x – 1).
Solution : By remainder theorem, the required remainder is equal to f(1). Now, f(x) = x3 – 3x2 + 2x + 1
Þ f(1) = 1 – 3 + 2 + 1 = 1.
Hence, the required remainder is equal to 1.
Factor theorem : Let f(x) be a polynomial of degree greater than or equal to one and a be a real number such that f(a) = 0, then (x – 1) is a factor of f(x). Conversely, if
(x – 1) is a factor of f(x), then f(a) = 0.
Remark : (1) (x + 1) is a factor of a polynomial f(x) if f(-a) = 0
- (ax – b) is a factor of a polynomial f(x) if f(b/a) = 0
- ax + b is a factor of a polynomial if + (- b/a) = 0
- (x – a) (x – b) is a factor of a polynomial f(x) if f(a) = 0 and f(b) =
Example : (x – 2) is a factor of the polynomial x3 – 3x2 – 4x + 12.
Solution : Let p(x) = x3 – 3×2 – 4x + 12 be the given polynomial. By factor theo- rem, (x – a) is a factor of a polynomial p(x) if p(a) = 0. Therefore, in order to prove that x
- 2 is a factor of p(x), it is sufficient to show that p(2) = 0. Now, p(x) = x3 – 3x2 – 4x + 12
p(2) = 23 – 3 C 22 – 4 C 2 + 12
= 8 – 12 – 8 + 12 = 0
Hence, (x – 2) is a factor of p(x) = x3 – 3x2 – 4x + 12
FACTORISATION TO OPEN PRODUCT INTO FACTORS
Taking out the common factors : When each term of an expression has a common factor, we divide each term by this factor and take it out as a multiple.
Grouping : The terms of the expression may be grouped so as to have a com- mon factor. It can now be factorised.
x2 ± 2xy + y2 = (x ± y)2.
Difference of two sequares : x2 – y2 = (x + y) (x – y)
Sum and difference of two cubes : x3 + y3 = (x + y) (x2 – xy + y2) x3 – y3 = (x – y) (x2 + xy + y2)
y2 + (a + b) y + ab = (y + a) (y +b) ay2 + by + c.
Factorisation of polynomials using remainder theorem : The following algo- rithm is useful in the factorization of polynomials over integers.
Algorithm :
Step 1. Write down the given polynomial f(x) (say).
Step 2. Get the constant term in f(x) and find its all possible factors.
For example, in the expression f(x) = x3 – 6x2 + 11x – 6 the constant term is – 6 and suppose its factors are ± 1, ± 2, ± 3, ± 6.
Step 3. Take one of the factors, say a1 and replace x by it in the given polyno- mial. If the polynomial reduces to zero, then x – a1 is a factor of the polynomial. Other-
wise, take another factor and continue this procedure till you get as many as factors as the degree of the polynomial.
Let x – a, x – a, x – a3,….. be the factors obtained by applying the above
pricedure.
Step 4. Put f(x) equal to k(x – a1) (x – a2) (x – a3). i.e., write f(x) = k(x – a1) (x – a2)
(x – a3)… , where k is a constant.
Step 5. Substitute any value of x other than a1, a2, a3,………………………………………………………….. in the equation obtained
in step 4 and get the value of k.
Step 6. Substitute the value of k in f(x) = k(x – a1) (x – a2) (x – a3)………………………………………………. Following
examples will illustrate the above procedure.
Example : Using factor theorem, factorize the polynomial x3 – 6×2 + 11x -6.
Solution : Let f(x) = x3 – 6x2 + 11x -6.
The constant term in f(x) is equal to -6 are ± 1, ± 2, ± 3, ± 6.
Putting x = 1 in f(x), we have f(1) = 13 – 6 C 12 + 11 C 1 – 6 = 1 – 6 + 11 – 6 = 0
\ (x – 1) is a factor of f(x).
Similarly, x – 2 and x – 3 are factor of f(x).
Since f(x) is an expression of degree 3. So, the polynomial cannot have more than three linear factors.
Let f(x) = k(x – 1) (x – 2) (x – 3) Then,
x3 – 6×2 + 11x – 6 = k(x – 1) (x – 2) (x – 3)
Putting x = 0 on both sides, we get – 6 = k(0 – 1) (0 – 2) (0 – 3) Þ – 6 = – 6k
Þ k = 1.
Putting k = 1 in f(x) = k(x – 1) (x – 2) (x – 3), we get
f(x) = (x – 1) (x – 2) (x – 3)
Hence, x3 – 6x2 + 11x – 6 = (x – 1) (x – 2) (x – 3).
Symmetric expressions : An expression having two or more letters is said to be symmetric if it remains unchanged when these letters are interchanged.
Example : The expression x2 + 5xy + y2 is symmetric with respect to x and y since when x and y are interchanged, the resulting expression y2 + 5yx + x2 is same as the original one.
Homogeneous expression : An expression in two or more variable is said to be a homogeneous of degree n, if the sum of the powers of all the variables in each term is equal to n.
Example : Clearly, the expression a + b + c is a homogeneous expression of degree one.
Principle of indeterminate coefficients : If two polynomials of the same degree in one variable are identically equal, then the coefficients of the like powers of the variable in the two polynomials are separately equal i.e.,
a + a x + a x2 + … + a xn = b + bx + b x2 + … + b xn
0 1 2 n 0 1 2 n
a0 = b0, a1 = b1, a2 = b2, …, an = b1.
Example : Using factor theorem we prove that
a2 (b – c) + b2 (c – a) + c2 (a – b) = – (a – b) (b – c) (c – a)
Solution : We have :
LHS = a2 (b – c) + b2 (c – a) + c2 (a – b) = a2b – a2c + b2c – b2a + c2a – c2b. Clearly, it is a homogeneous expression of degree 3.
Putting a = b in the given expression it becomes
= b2. b – b2c + b2c – b2. b + c2b – c2b = 0
So, (a – b) is a factor of the given expression.
Similarly, by symmetry each of (b – c) and (c – a) is a factor of the expression.
Now, since the given expression is a symmetrical and homogeneous expression of degree 3. So, it cannot have more than 3 linear factors.
Therefore, let
a2(b – c) + a2 (c – a) + c2 (a – b) º k (a – b) (b – c) (c – a) where k is a constant.
Now, putting any convenient values of a, b, c in the above identify such that neither side of (1) becomes zero, we may obtain the value of k.
Putting a = 0, b = 1 and c = – 1 in (1) we get
0 + 1 (-1 – 0) + (-1)2 (0 – 1) = k(0 – 1) (1 + 1) (- 1 – 0)
Þ -2 = 2k Þ k = -1 Substituting k = – 1
a2 (b – c) + b2 (c – a) + c2 (a – b) = – (b – b) (b – c) (c – a).
Properties of Properties of
Integers Polynomials
- The sum of two The sum of two
integers is an polynomials is a
integer. polynomial.
- Addition of Addition of integers is commutative. polynomials is
commutative.
- Addition of integer is Addition of polynomials is associative,
associative, i.e., (a + b) i.e., p(x) + {q(x) + r(x)} = {p(x) + q(x)}
+ c = a + (b + c) for all + r(x) for all polynomials p(x), r(x). integers a, b, c.
- The integer (0). is such 4. The zero polynomials (0), is such that that a + 0 = a = 0 + a for any p(x) + 0 = p(x) = 0 + p(x) for any integer a. polynomial p(x).
- For any integer a, there For any polynomial p(x), there corresponds an integer corresponds a polynomial -p(x) such
– a such that a + (-a) = that p(x) + [- p(x)] = 0 = [- p (x)] +p(x). 0 = (-a) + a.
- The product of any two The product of any two polynomials is a integers is an integer. polynomial.
- Multiplication of integers is Multiplication of polynomials is commutative i.e., a.b = b.a commutative i.e. p(x).q(x) = q(x). p(x) for for any two integers a, b. any two polynomials p(x) q(x).
- Multiplication of integers is Multiplication of polynomials is associative, i.e., (a.b).c = associative i.e., {p(x).q(x)} r(x) = p(x).
- (b.c) for all integers a, b, {q(x).r(x)} for all polynomials p(x), q(x),
r(x).
- Multiplication of integers is Multiplication of polynomials is distributive over addition, i.e., distributive over addition, i.e., p(x).{q(x) a. (b + c) = a.b + a.c(b + c) +r(x)} = p(x).q(x) + p(x).r(x), (q(x) + r(x)). a = b.a + c.a for all integers p(x) q(x).p(x) + r(x). p(x) for all
a, b, c. polynomial p(x), q(x), r(x).
- The integer l is such that The constant polynomial l is such that
- l = a = a for any integer a. p(x). l = p(x) = l. p(x) for any polynomial
p(x).
Thus, we may say that polynomials behave like integers.
PASCAL’S Triangle : This triangle is used to get coefficients e.g
1 n = 0
1 1 = 1
1 2 1 = 2
1 3 3 1 = 3
1 4 6 4 1 = 4
1 5 10 10 5 1 = 5
1 6 15 20 15 6 1 = 6
1 7 21 35 35 21 7 1 = 7
1 8 28 56 70 56 28 8 1 = 8
