Polynomials are one of the key sections in the new CBSE Class 9 Maths syllabus. It helps develop an understanding of algebraic expressions and their operations. The chapter introduces the student to the definition of various types of polynomials as well as the operations related to polynomials. By properly defining the terms, recording formulae, offering solved examples, and providing revision advice, these notes truly seek to help students understand the ideas and provide them with a clear roadmap for solving maths problems. By using these notes, students may improve their problem-solving abilities and achieve high test scores.
Polynomial Definition: A polynomial is an algebraic expression consisting of variables, coefficients, and non-negative integer exponents.
Types of Polynomials:
Degree and Zeros of a Polynomial
Dividend = (Divisor × Quotient) + Remainder
Or,
p (x) = (x−a) q (x)+ r
Here, q (x) is the quotient, and r is the remainder.
For the division of polynomials, follow the following steps
Step 1: Write the terms of both polynomials in descending order of degrees (if they are not already in that order).
Step 2: Divide the first term of the dividend (the polynomial you are dividing) by the first term of the divisor (the polynomial you are dividing by).
Step 3: Multiply the whole divisor by the term you just found (from step 2).
Step 4: Subtract the result obtained in Step 3 from the dividend.
Step 5: Now repeat the division with a new polynomial.
Step 6: Subtract again
Step 7: Check whether the degree of the remainder is less than the degree of the divisor.
Example 1: Divide the polynomial P(x) = 2x3−3x2+4x−5 by (x−2). Find the remainder using the Remainder Theorem.
Solution: give that x-2 is divisor
x - 2 = 0; x = 2
P(x) = 2x3 − 3x2 + 4x − 5 put x = 2
P(2) = 2 x 23−3 x 22+ 4 x 2 − 5
= 16 - 12 + 8 - 5 = 7
Example 2: Determine if (x−2) is a factor of the polynomial P(x) = x3 − 4x2 + 3x + 2.
Solution: x-2 = 0; x = 2
Put x = 2 in P(x) = x3 − 4x2 + 3x + 2 to check whether (x-2) is factor or not.
P(2) = 23 − 422 + 3 x 2 + 2
P(2) = 8 – 16 + 6 + 2
= 16 – 16 = 0
Since p(2) = 0, hence x-2 is a factor of the given equation.
Example 3: Find the remainder when is divided by 𝑥−2.
Solution:
Using the Remainder Theorem, substitute x=2 into p (x):
The remainder is 6.
Example 4: Divide 3x4 + x3 - 17x2 + 19x - 6 by 3x2 + 7x - 6.
Solution:
Here, in this example, the remainder is 0.
Example 5: Find the value of ‘a’ if x – a is a factor of x3 – ax2 + 2x + a – 1.
Solution:
Let p(x) = x3 – ax2 + 2x + a – 1
Since (x – a) is a factor of p(x), p(a) = 0.
x - a = 0; x = a
p(a) = a3 – a(a)2 + 2a + a – 1 = 0
a3 – a3 + 2a + a – 1 = 0
3a = 1
Hence, a = 1/3
Example 6: If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution:
Using identity
= (x+y)2 = x2 + 2xy + y2
= 122 = x2 + y2 + 227
= x2 + y2 = 144 - 54 = 90
Now, x3 + y3 = (x + y) (x2 + y2– xy)
= 12 [90 - 27]
= 12 × 63 = 756
Example 7: Factorise: 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz
Solution:
Using the identity
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
We can also write this equation 9x2 + 4y2 + 16z2 + 12xy – 16yz – 24xz as
= (3x+2y-4z)2 = (3x+2y-4z)(3x+2y-4z)
(Session 2025 - 26)