NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.2 Factorisation
NCERT Solutions for Class 8 Maths Chapter 12 – Factorisation Ex 12.2 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. To factorise means to write an expression as a product of its factors. Exercise 12.1 taught you how to take out a common factor. Exercise 12.2 teaches a faster way, using standard identities. An identity is a formula that is always true, whatever numbers you put in.
This exercise has 5 questions with many parts, so there is a lot of practice here. You will use the square identities, the difference of two squares, and the method of regrouping terms in pairs. Every part below is solved in short steps, with the final factors in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.
NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.2 Factorisation
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Exercise 12.2 – Questions and Answers
Q 1. Factorise the following expressions.
(i) a² + 8a + 16
(ii) p² − 10p + 25
(iii) 25m² + 30m + 9
(iv) 49y² + 84yz + 36z²
(v) 4x² − 8x + 4
(vi) 121b² − 88bc + 16c²
(vii) (l + m)² − 4lm (Hint: Expand (l + m)² first)
(viii) a⁴ + 2a²b² + b⁴
Answer:
(i) a² + 8a + 16
The given expression can be written as:
= (a)² + 2 × a × 4 + (4)²
This is of the form x² + 2xy + y² with x = a, y = 4 and 2xy = 8a
Therefore, by using the identity (x + y)² = x² + 2xy + y², we get
= (a + 4)²
(ii) p² − 10p + 25
The given expression can be written as:
= (p)² − 2 × p × 5 + (5)²
This is of the form a² − 2ab + b² with a = p, b = 5 and 2ab = 10p
Therefore, by using the identity (a − b)² = a² − 2ab + b², we get
= (p − 5)²
(iii) 25m² + 30m + 9
The given expression can be written as:
= (5m)² + 2 × 5m × 3 + (3)²
This is of the form a² + 2ab + b² with a = 5m, b = 3 and 2ab = 30m
Therefore, by using the identity (a + b)² = a² + 2ab + b², we get
= (5m + 3)²
(iv) 49y² + 84yz + 36z²
The given expression can be written as:
= (7y)² + 2 × (7y) × (6z) + (6z)²
This is of the form a² + 2ab + b² with a = 7y, b = 6z and 2ab = 84yz
Therefore, by using the identity (a + b)² = a² + 2ab + b², we get
= (7y + 6z)²
(v) 4x² − 8x + 4
The given expression can be written as:
= (2x)² − 2 (2x) (2) + (2)²
This is of the form a² − 2ab + b² with a = 2x, b = 2 and 2ab = 8x
Therefore, by using the identity (a − b)² = a² − 2ab + b², we get
= (2x − 2)²
Again, we can take out 2 common
= [(2) (x − 1)]²
= 4(x − 1)²
(vi) 121b² − 88bc + 16c²
The given expression can be written as:
= (11b)² − 2 (11b) (4c) + (4c)²
This is of the form a² − 2ab + b² with a = 11b, b = 4c and 2ab = 88bc
Therefore, by using the identity (a − b)² = a² − 2ab + b², we get
= (11b − 4c)²
(vii) (l + m)² − 4lm
The given expression can be written as:
= l² + 2lm + m² − 4lm
= l² − 2lm + m²
This is of the form a² − 2ab + b² with a = l, b = m and 2ab = 2lm
Therefore, by using the identity (a − b)² = a² − 2ab + b², we get
= (l − m)²
(viii) a⁴ + 2a²b² + b⁴
The given expression can be written as:
= (a²)² + 2 (a²) (b²) + (b²)²
This is of the form x² + 2xy + y² with x = a², y = b² and 2xy = 2a²b²
Therefore, by using the identity (x + y)² = x² + 2xy + y², we get
= (a² + b²)²
Q 2. Factorise.
(i) 4p² − 9q²
(ii) 63a² − 112b²
(iii) 49x² − 36
(iv) 16x⁵ − 144x³
(v) (l + m)² − (l − m)²
(vi) 9x²y² − 16
(vii) (x² − 2xy + y²) − z²
(viii) 25a² − 4b² + 28bc − 49c²
Answer:
(i) 4p² − 9q²
The given expression can be written as:
= (2p)² − (3q)²
This is of the form a² − b² with a = 2p and b = 3q
Therefore, by using the identity a² − b² = (a − b)(a + b), we get
= (2p + 3q) (2p − 3q)
(ii) 63a² − 112b²
The given expression can be written as:
= 7(9a² − 16b²)
= 7[(3a)² − (4b)²]
This is of the form x² − y² with x = 3a and y = 4b
Therefore, by using the identity x² − y² = (x − y)(x + y), we get
= 7(3a + 4b) (3a − 4b)
(iii) 49x² − 36
The given expression can be written as:
= (7x)² − (6)²
= (7x − 6) (7x + 6)
(iv) 16x⁵ − 144x³
The given expression can be written as:
= 16x³(x² − 9)
= 16x³ [(x)² − (3)²]
This is of the form a² − b² with a = x and b = 3
Therefore, by using the identity a² − b² = (a − b)(a + b), we get
= 16x³(x − 3) (x + 3)
(v) (l + m)² − (l − m)²
This is of the form a² − b² with a = (l + m) and b = (l − m)
Therefore, by using the identity a² − b² = (a − b)(a + b), we get
= [(l + m) − (l − m)] [(l + m) + (l − m)]
= (l + m − l + m) (l + m + l − m)
= 2m × 2l
= 4ml
= 4lm
(vi) 9x²y² − 16
The given expression can be written as:
= (3xy)² − (4)²
This is of the form a² − b² with a = 3xy and b = 4
Therefore, by using the identity a² − b² = (a − b)(a + b), we get
= (3xy − 4) (3xy + 4)
(vii) (x² − 2xy + y²) − z²
By using the identity (a − b)² = a² − 2ab + b², we get
= (x − y)² − (z)²
This is of the form a² − b² with a = (x − y) and b = z
Therefore, by using the identity a² − b² = (a − b)(a + b), we get
= (x − y − z) (x − y + z)
(viii) 25a² − 4b² + 28bc − 49c²
The given expression can be written as:
= 25a² − (4b² − 28bc + 49c²)
= (5a)² − [(2b)² − 2 × 2b × 7c + (7c)²]
By using the identity (a − b)² = a² − 2ab + b², we get
= (5a)² − [(2b − 7c)²]
This is of the form x² − y² with x = 5a and y = (2b − 7c)
Therefore, by using the identity x² − y² = (x − y)(x + y), we get
= [5a + (2b − 7c)] [5a − (2b − 7c)]
= (5a + 2b − 7c) (5a − 2b + 7c)
Q 3. Factorise the expressions.
(i) ax² + bx
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + l) + m + 1
(vi) y (y + z) + 9 (y + z)
(vii) 5y² − 20y − 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy − 4y + 6 − 9x
Answer:
(i) ax² + bx
= a × x × x + b × x
Here, the common factor is 'x'.
Therefore, ax² + bx = x(ax + b)
(ii) 7p² + 21q²
= 7 × p × p + 3 × 7 × q × q
Here, the common factor is '7'.
Therefore, 7p² + 21q² = 7(p² + 3q²)
(iii) 2x³ + 2xy² + 2xz²
Here, the common factor is '2x'.
Therefore, 2x³ + 2xy² + 2xz² = 2x(x² + y² + z²)
(iv) am² + bm² + bn² + an²
The given expression can be written as:
= am² + bm² + an² + bn²
The common factor in the first two terms is m² and the common factor in the last two terms is n².
= m²(a + b) + n²(a + b)
= (a + b) (m² + n²)
Therefore, am² + bm² + bn² + an² = (a + b) (m² + n²)
(v) (lm + l) + m + 1
Rearranging the given expression as follows:
= lm + m + l + 1
Taking out common factors:
= m(l + 1) + 1(l + 1)
= (l + 1) (m + 1)
Therefore, (lm + l) + m + 1 = (l + 1) (m + 1)
(vi) y (y + z) + 9 (y + z)
= (y + z) (y + 9)
Therefore, y (y + z) + 9 (y + z) = (y + z) (y + 9)
(vii) 5y² − 20y − 8z + 2yz
The given expression can be written as follows:
= 5y² − 20y + 2yz − 8z
Taking out common factors:
= 5y(y − 4) + 2z(y − 4)
= (y − 4) (5y + 2z)
Therefore, 5y² − 20y − 8z + 2yz = (y − 4) (5y + 2z)
(viii) 10ab + 4a + 5b + 2
Rearranging the given expression as follows:
= 10ab + 5b + 4a + 2
Taking out the common terms:
= 5b(2a + 1) + 2(2a + 1)
= (2a + 1) (5b + 2)
Therefore, 10ab + 4a + 5b + 2 = (2a + 1) (5b + 2)
(ix) 6xy − 4y + 6 − 9x
Rearranging the terms of the given expression and taking out common terms:
= 6xy − 9x − 4y + 6
= 3x(2y − 3) − 2(2y − 3)
= (2y − 3) (3x − 2)
Therefore, 6xy − 4y + 6 − 9x = (2y − 3) (3x − 2)
Q 4. Factorise the following expressions.
(i) p² + 6p + 8
(ii) q² − 10q + 21
(iii) p² + 6p − 16
Answer:
(i) p² + 6p + 8
Here, 8 = 4 × 2 and 6 = 4 + 2
Therefore, p² + 6p + 8 can be written as:
p² + 2p + 4p + 8
= p(p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q² − 10q + 21
Here, 21 = (−7) × (−3) and −10 = (−7) + (−3)
Therefore, q² − 10q + 21 can be written as:
q² − 7q − 3q + 21
= q(q − 7) − 3(q − 7)
= (q − 7) (q − 3)
(iii) p² + 6p − 16
Here, −16 = (−2) × 8 and 6 = 8 + (−2)
Therefore, p² + 6p − 16 can be written as:
= p² + 8p − 2p − 16
= p(p + 8) − 2(p + 8)
= (p + 8) (p − 2)
Q 5. Factorise.
(i) a⁴ − b⁴
(ii) p⁴ − 81
(iii) x⁴ − (y + z)⁴
(iv) x⁴ − (x − z)⁴
(v) a⁴ − 2a²b² + b⁴
Answer:
(i) a⁴ − b⁴
= (a²)² − (b²)²
= (a² − b²) (a² + b²) [By using a² − b² = (a + b)(a − b)]
= (a − b) (a + b) (a² + b²) [By using a² − b² = (a + b)(a − b)]
(ii) p⁴ − 81
= (p²)² − (9)²
= (p² − 9) (p² + 9) [By using a² − b² = (a + b)(a − b)]
= [(p)² − (3)²] (p² + 9)
= (p − 3) (p + 3) (p² + 9) [By using a² − b² = (a + b)(a − b)]
(iii) x⁴ − (y + z)⁴
= (x²)² − [(y + z)²]²
= [x² − (y + z)²] [x² + (y + z)²] [By using a² − b² = (a + b)(a − b)]
= [x − (y + z)] [x + (y + z)] [x² + (y + z)²] [By using a² − b² = (a + b)(a − b)]
= (x − y − z) (x + y + z) [x² + (y + z)²]
(iv) x⁴ − (x − z)⁴
= (x²)² − [(x − z)²]²
= [x² − (x − z)²] [x² + (x − z)²] [By using a² − b² = (a + b)(a − b)]
= [x − (x − z)] [x + (x − z)] [x² + (x − z)²] [By using a² − b² = (a + b)(a − b)]
= z (2x − z) [x² + x² − 2xz + z²]
= z (2x − z) (2x² − 2xz + z²)
(v) a⁴ − 2a²b² + b⁴
= (a²)² − 2 (a²) (b²) + (b²)²
= (a² − b²)²
= [(a − b) (a + b)]² [By using a² − b² = (a + b)(a − b)]
= (a − b)² (a + b)²
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