Find the best tutors and institutes for Class 10 Tuition
Q3(ii):
Factorise the following using appropriate identities:
(ii) $4y^2 – 4y + 1$
Solution :
Step 1: Initial Expression Analysis
We are given the quadratic trinomial:
$P(y) = 4y^2 - 4y + 1$
Our objective is to factorise this expression by identifying it as a perfect square. A quadratic trinomial takes the general form $ax^2 + bx + c$. We must analyze the first and last terms to determine if they are perfect squares.
Step 2: Identifying the Appropriate Algebraic Identity
We observe that the signs of the terms alternate ($+, -, +$). This suggests the use of the standard binomial square identity for subtraction [Per the fundamental axioms of polynomial expansion]:
$(a - b)^2 = a^2 - 2ab + b^2$
Step 3: Term-by-Term Transformation & Verification
We must map the terms of our given polynomial to the components of the identity $a^2 - 2ab + b^2$.
- First Term ($a^2$): The term $4y^2$ can be rewritten as the square of a single monomial.
$4y^2 = (2y)^2 \implies a = 2y$ - Third Term ($b^2$): The constant $1$ is a perfect square.
$1 = (1)^2 \implies b = 1$ - Middle Term ($-2ab$): We must verify that the middle term of our polynomial matches $-2ab$ using our derived values for $a$ and $b$.
$-2ab = -2(2y)(1) = -4y$
Since the middle term perfectly matches the given expression ($-4y$), the polynomial is confirmed to be a perfect square trinomial.
Step 4: Applying the Identity
By substituting $a = 2y$ and $b = 1$ back into the structural framework of the identity, we can rewrite the expanded polynomial as a squared binomial:
$4y^2 - 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2$
$4y^2 - 4y + 1 = (2y - 1)^2$
Step 5: Geometric Verification (Area Model)
To rigorously prove this factorization, we can use an area model. Consider a large square with a total side length of $2y$. Its total area is $(2y)^2 = 4y^2$. If we partition this square into segments of length $(2y - 1)$ and $1$, we can geometrically derive the identity.
From the geometric model, the total area of the square is the sum of its four internal regions:
$Area_{Total} = (2y - 1)^2 + (2y - 1) + (2y - 1) + 1$
$4y^2 = (2y - 1)^2 + 4y - 2 + 1$
$4y^2 = (2y - 1)^2 + 4y - 1$
Isolating the primary region $(2y - 1)^2$ yields:
$(2y - 1)^2 = 4y^2 - 4y + 1$
[This confirms our algebraic factorization is geometrically absolute].
Step 6: Final Factorisation
To express the polynomial fully in its factorised form, we write the squared binomial as the product of two identical linear binomials.
$(2y - 1)^2 = (2y - 1)(2y - 1)$
Final Solution: The factorised form of the polynomial $4y^2 - 4y + 1$ is $(2y - 1)(2y - 1)$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
Top Tutors who teach Polynomials
I am R.Vinotha ., M.Sc.,M.Phil.,B.Ed in mathematics.I am the trainer/tutor of this course .I was allotted 12 regular classes for a month.I am using through google meet app for online class and also i am taking mathematics tuition at my home.I will create seperate WhatApp group for my tuition students and communicates to them .I will send the class notes to them after completing the class .My class planner is us following: Regular classes : Any 3 days for a week Class duration : 1 hour Home works: Daily write two problems and send it to be in WhatsApp group Test: Daily test : Formulae test: Everyday i am asking 10 formulae through voice message in WhatsApp group after getting my message ,the students will answer the questions and send it tome via tuition group Problems test:Daily write two problems Regular test : Every saturday and sunday Monthly test : Every month fourth saturday doubts clarification time : Every day 7 pm -8 pm I will correct the test papers immedietly and send it to their personal windows and also resolve the students mistakes. Chapter's quiz : After completing every chapters ,i will conduct mathematical quiz for that chapters Revisions: I will revise previous years questio papers After this type of my coachings many of my students got 100 marks in theire board examination .I put lots of efforts to my students and they getting good marks in the board examination .My teaching type is friendly manner.I gave extra solved questions and answers to my students .
Taking class as well as gives coaching in different approaches that can be dealt with the topic thro variety in problems and helps to reach solution with clarity of the nuances to be careful with.
I have specifically guided Class 10 students for board exams, with a strong focus on NCERT syllabus, and topics like Real Numbers, Polynomials, Quadratic Equations, and Statistics. I provide regular tests, personalized feedback, and exam-oriented strategies to improve accuracy and speed. Many of my students have shown remarkable improvement and scored above 90% in their board exams.
It was an amazing experience learning and understanding maths with Saurav sir. He really helped me overcome my fear of maths and helped me score good marks in a subject that previously scared me. It really shows how dedicated he is as a teacher. Really good experience!
Teaching is my passion I am into teaching profession since 16 years. I like challenges with young minds.. I will make students to confident by explaining concepts, taking special care. Cent percent assurance to child growth, committed to work towards target. I used to counsel the students to succeed in their exams.
Experienced Mathematics and Science Educator | 10+ Years in Teaching and Coaching I have over a decade of experience in teaching and coaching students in Mathematics and Science. I have worked as a PGT Mathematics teacher at GBL Convent School, and have also taught Classes 9 and 10 (Maths & Science) at R.S. Education Hub. My teaching approach focuses on building strong conceptual clarity, personalized attention, and result-oriented strategies to help students excel academically. I have successfully guided numerous students through school exams, board exams, and foundational competitive preparation.
Coaching is about improvising a student from where he/she is at. Not about converting an already bright student into an achiever. It should be obvious. The whole idea of ONE ON ONE coaching is to improvise and to elimate the problems in a detailed manner, exactly the way a student desires. To connect and respond to a student cannot be everybodys cup of tea. This is where a capable one comes in to fill the gaps where modern day educationalists fail to fill.
Find more Tutor for Polynomials in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
Download free CBSE - Class 9 Mathematics Polynomials EXERCISE 2.4 worksheets
Download Now