Notes of Polynomials
Topics in the Chapter- Polynomials
- Classification of polynomials on the basis of number of terms
- Degree of a Polynomial
- Values of polynomials at different points
- Zeroes of a polynomial
- Identity: (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
- Remainder Theorem
- Factor Theorem
- Identity: x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
Polynomials
→ An algebraic expression in which the exponents of the variables are non-negative integers
are called polynomials.
For example: 3x4 + 2x3 + x + 9, 3x2 etc.
Constant polynomial
→ A constant polynomial is of the form p(x) = k, where k is a real number.
For example, –9, 10, 0 are constant polynomials.
Zero polynomial
A constant polynomial ‘0’ is called zero polynomial.
General form of a polynomial
A polynomial of the form where are p(x) = anxn + an-1xn-1 + .... + a1x + a0x where a0, a1... ar, are constant and an≠ 0.
Here, a0, a1... an are the respective coefficients of x0,x1, x2 ... xn and n is the power of
the variable x.
anxn + an-1xn-1 - a0 and are called the terms of p(x).
Classification of polynomials on the basis of number of terms
• A polynomial having one term is called a monomial. e.g. 3x, 25t3 etc.
• A polynomial having one term is called a monomial. e.g. 2t-6, 3x4 +2x etc.
• A polynomial having one term is called a monomial. e.g. 3x4 + 8x + 7 etc.
Degree of a Polynomial
• The degree of a polynomial is the highest exponent of the variable of the polynomial.
For example, the degree of polynomial 3x4 + 2x3 + x + 9 is 4.
• The degree of a term of a polynomial is the value of the exponent of the term.
Classification of polynomial according to their degrees
• A polynomial of degree one is called a linear polynomial e.g. 3x+2, 4x, x+9.
• A polynomial of degree is called a quadratic polynomial. e.g. x2+ 9, 3x2 + 4x + 6
• A polynomial of degree three is called a cubic polynomial e.g. 10x3+ 3, 9x3
Note: The degree of a non-zero constant polynomial is zero and the degree of a zero polynomial is not defined.
Values of polynomials at different points
• A polynomial is made up of constants and variables. Hence, the value of the polynomial changes
as the value of the variable in the polynomial changes.
• Thus, for the different values of the variable x, we get different values of the polynomial.
Zeroes of a polynomial
• A real number a is said to be the zero of polynomial p(x) if p(a) = 0, In this case, a is also called the root of the equation p(x) = 0.
Note:
• The maximum number of roots of a polynomial is less than or equal to the degree of the polynomial.
• A non-zero constant polynomial has no zeroes.
• A polynomial can have more than one zero.
Steps of the Division of a Polynomial with a Non –Zero Polynomial
Divide x2 - 3x -10 by 2 + x
Step 1: Write the dividend and divisor in the descending order i.e. in the standard form. x2 - 3x -10 and x + 2
Divide the first term of the dividend with the first term of the divisor.
x2/x = x this will be the first term of the quotient.
Step 2: Now multiply the divisor with this term of the quotient and subtract it from the dividend.
Step 3: Now the remainder is our new dividend so we will repeat the process again by dividing the dividend with the divisor.
Step 4: – (5x/x) = – 5
Step 5:
The remainder is zero.
Hence x2 - 3x – 10 = (x + 2)(x - 5) + 0
Dividend = (Divisor × Quotient) + Remainder
Remainder Theorem says that if p(x) is any polynomial of degree greater than or equal to one and let ‘t’ be any real number and p (x) is divided by the linear polynomial x – t, then the remainder is p(t).
As we know that
P(x) = g(x) q(x) + r(x)
If p(x) is divided by (x-t) then
If x = t
P (t) = (t - t).q (t) + r = 0
To find the remainder or to check the multiple of the polynomial we can use the remainder theorem.
Example:
What is the remainder if a4 + a3 – 2a2 + a + 1 is divided by a – 1.
Solution:
P(x) = a4 + a3 – 2a2 + a + 1
To find the zero of the (a – 1) we need to equate it to zero.
a -1 = 0
a = 1
p (1) = (1)4 + (1)3 – 2(1)2 + (1) + 1
= 1 + 1 – 2 + 1 + 1
= 2
So by using the remainder theorem, we can easily find the remainder after the division of polynomial.
Factor Theorem
Factor theorem says that if p(y) is a polynomial with degree n≥1 and t is a real number, then
-
(y - t) is a factor of p(y), if p(t) = 0, and
-
P (t) = 0 if (y – t) is a factor of p(y).
Example: 1
Check whether g(x) = x – 3 is the factor of p(x) = x3 - 4x2 + x + 6 using factor theorem.
Solution:
According to the factor theorem if x - 3 is the factor of p(x) then p(3) = 0, as the root of x – 3 is 3.
P (3) = (3)3 - 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6 = 0
Hence, g(x) is the factor of p(x).
Example: 2
Find the value of k, if x – 1 is a factor of p(x) = kx2 – √2x + 1
Solution:
As x -1 is the factor so p(1) = 0
Factorization of Polynomials
Factorization can be done by three methods
1. By taking out the common factor
If we have to factorize x2 –x then we can do it by taking x common.
x(x – 1) so that x and x-1 are the factors of x2 – x.
2. By grouping
ab + bc + ax + cx = (ab + bc) + (ax + cx)
= b(a + c) + x(a + c)
= (a + c)(b + x)
3. By splitting the middle term
x2 + bx + c = x2 + (p + q) + pq
= (x + p)(x + q)
This shows that we have to split the middle term in such a way that the sum of the two terms is equal to ‘b’ and the product is equal to ‘c’.
Example: 1
Factorize 6x2 + 17x + 5 by splitting the middle term.
Solution:
If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.
Some of the factors of 30 are 1 and 30, 2 and 15, 3 and 10, 5 and 6, out of which 2 and 15 is the pair which gives p + q = 17.
6x2 + 17x + 5 =6 x2 + (2 + 15) x + 5
= 6 x2 + 2x + 15x + 5
= 2 x (3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
| Algebraic Identities |
| 1. (x + y)2 = x2 + 2xy + y2 |
| 2. (x - y)2 = x2 - 2xy + y2 |
| 3. (x + y) (x - y) = x2 - y2 |
| 4. (x + a) (x + b) = x2 + (a + b)x + ab |
| 5. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx |
| 6. (x + y)3 = x3 + y3 + 3xy(x + y) = x3+ y3 + 3x2y + 3xy2 |
| 7. (x - y)3 = x3- y3 - 3xy(x - y) = x3 - y3 - 3x2y + 3xy2 |
| 8. x3 + y3 = (x + y)(x2 – xy + y2) |
| 9. x3 - y3 = (x - y)(x2 + xy + y2) |
| 10. x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz - zx) x3 + y3 + z3 = 3xyz if x + y + z = 0 |
Example: 2
Factorize 8x3 + 27y3 + 36x2y + 54xy2
Solution:
The given expression can be written as
= (2x)3 + (3y)3 + 3(4x2) (3y) + 3(2x) (9y2)
= (2x)3 + (3y)3 + 3(2x)2(3y) + 3(2x)(3y)2
= (2x + 3y)3 (Using Identity VI)
= (2x + 3y) (2x + 3y) (2x + 3y) are the factors.
Example: 3
Factorize 4x2 + y2 + z2 – 4xy – 2yz + 4xz.
Solution:
4x2 + y2 + z2 – 4xy – 2yz + 4xz = (2x)2 + (–y)2 + (z)2 + 2(2x) (-y)+ 2(–y)(z) + 2(2x)(z)
= [2x + (- y) + z]2 (Using Identity V)
= (2x – y + z)2 = (2x – y + z) (2x – y + z)
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