NCERT Revision Notes for Chapter 2 Polynomials Class 9 Maths
CBSE NCERT Revision Notes1
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An algebraic expression in which the exponents of the variables are non-negative integers are called polynomials.
For example, 3x4 + 2x3 + x + 9, 3x4 etc are polynomials.
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.
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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).
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• 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.
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• 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.
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• 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.
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Example: Divide x4 -2x3- 2x2+7x-15 by x-2
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Other ways to represent these identities are:
x3+ y3 = (x+y)3 - 3xy(x+y)
x3+ y3 = (x+y)(x2-xy+y2)
x3- y3 = (x-y)3 + 3xy(x-y)
x3- y3 = (x-y)(x2+xy+y2)
Example: Expand (3x+2y)3 - (3x-2y)3
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If p(x) is a polynomial of degree greater than or equal to one and a is any real number then if p(x) is divided by the linear polynomial x-a, the remainder is p(a).
Example:
Find the remainder when x5 -x2 +5 is divided by x-2.
Solution:
p(x) = x5 -x2 +5
The zero of x-2 is 2.
p(2) = 25 -22 +5 = 32-4+5 = 33
Therefore, by remainder theorem, the remainder is 33.
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If p(x) is a polynomial of degree n≥1 and a is any real number, then
x-a is a factor of p(x), if p(a)=0.
p(a) = 0, if (x-a) is a factor of p(x).
Example:
Determine whether x +3 is a factor of x3+5x2+5x-3.
Solution
The zero of x+3 is -3.
Let p(x) = x3+5x2+5x-3
p(-3) = (3)3+5(-3)2 + 5(-3)-3
= -27+45-15-3
= -45 + 45
= 0
Therefore, by factor theorem, x + 3 is the factor of p(x).
Factorisation of quadratic polynomials of the form ax2+bx+c can be done using Factor
theorem and splitting the middle term.
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Let p(x) = x2– 7x + 10
The constant term is 0 and its factors are ±1, ±2, ±5 and ±1O.
Let us check the value of the polynomial for each of these factors of 10.
p(1) = 12– 7×1 + 10 = 1-7+10 = 4 ≠ 0
Hence, x-1 is not a factor of p(x),
p(2) = 22– 7×2 + 10 = 4-14+10 = 0
Hence, x -2 is a factor of p(x).
p(5) = 52- 7×5 +10 = 25-35+10 = 0
Hence, x -5 is a factor of p(x).
We know that a quadratic polynomial can have a maximum of two factors. We have obtained the two factors of the given polynomial, which are x-2 and x-5.
Thus, we can write the given polynomial as:
Let p(x) = x2– 7x + 10 = (x-2) (x-5)
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The given polynomial is 2x2– 11x + 15
Here, a = 2×15 = 30. The middle term is -11. Therefore, we have split -11 into two numbers such that their product is 30 and their sum is -11. These numbers are -5 and -6 [As (-5)+ (-6) = -11 and -5 × -6 = 30]
Thus, we have:
2x2– 11x + 15 = 2x2– 5x - 6x + 15
= x(2x- 5) -3 (2x- 5)
= (x-3) (2x-5)
Note: A quadratic polynomial can have a maximum of two factors.
• Factorisation of cubic polynomials of the form ax3+ bx2+ cx can be done using factor
theorem and hit and trial method.
A cubic polynomial can have a maximum of three linear factors. So, by knowing one of these factors, we can reduce it to a quadratic polynomial.
Thus, to factorize a cubic polynomial, we first find a factor by the hit and trial method or by using the factor theorem, and then reduce the cubic polynomial into a quadratic polynomial and it is then solved further.
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The constant term is 6.
The factors of 6 are ±1, ±2, ±3 and ±6.
Let x = 1
From equation (1), we get
p(x) = (x – 1) (x – 2) (x + 3)
Hence, factors of polynomial p(x) are (x – 1), (x – 2) and (x + 3).
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