Class 8 Factorisation Ex 12.2

Keeping the examination point of view in mind the skmath.in team has prepared NCERT Solutions for Class-8 Maths- Factorisation Ex 12.2. The NCERT Solutions for chapter Factorisation solutions explains the easy and simple way to solve the problems. By understanding these ways in NCERT Solutions for Class 8, students will be confident while solving such problems found in Class-8 Maths- Factorisation Ex 12.2.

Question 1: Factorise the following expressions:

(i) \(a^2 + 8a + 16\)

Solution:

\(a^2 + 8a + 16 = (a)^2 + 2 × a × 4 + (4)^2\)

\(= (a + 4)^2 \ \ \ \ \ \ \ [\because (x + y)^2 = x^2 + 2xy + y^2]\)

(ii) \(p^2 − 10p + 25\)

Solution:

\(p^2 − 10p + 25 = (p)^2 − 2 × p × 5 + (5)^2\)

\(= (p − 5)^2 \ \ \ \ \ \ \ [\because(a − b)^2 = a^2 − 2ab + b^2]\)

(iii) \(25m^2 + 30m + 9\)

Solution:

\(25m^2 + 30m + 9 = (5m)^2 + 2 × 5m × 3 + (3)^2\)

\(= (5m + 3)^2 \ \ \ \ \ \ \ [\because(a + b)^2 = a^2 + 2ab + b^2]\)

(iv) \(49y^2 + 84yz + 36z^2\)

Solution:

\(49y^2 + 84yz + 36z^2 = (7y)^2 + 2 × (7y) × (6z) + (6z)^2\)

\(= (7y + 6z)^2 \ \ \ \ \ \ \ [\because(a + b)^2 = a^2 + 2ab + b^2]\)

(v) \(4x^2 − 8x + 4\)

Solution:

\(4x^2 − 8x + 4 = (2x)^2 − 2 (2x) (2) + (2)^2\)

\(= (2x − 2)^2 \ \ \ \ \ \ \ [\because(a − b)^2 = a^2 − 2ab + b^2]\)

\(= [(2) (x − 1)]^2 = 4(x − 1)^2\)

(vi) \(121b^2 − 88bc + 16c^2\)

Solution:

\(121b^2 − 88bc + 16c^2 = (11b)^2 − 2 (11b) (4c) + (4c)^2\)

\(= (11b − 4c)^2 \ \ \ \ \ \ \ [\because(a − b)^2 = a^2 − 2ab + b^2]\)

(vii) \((l + m)^2 − 4lm \)(Hint: Expand \((l + m)^2\) first)

Solution:

\((l + m)^2 − 4lm = l^2 + 2lm + m^2 − 4lm\)

\(= l^2 − 2lm + m^2\)

\(= (l − m)^2 \ \ \ \ \ \ \ [\because(a − b)^2 = a^2 − 2ab + b^2]\)

(viii) \(a^4 + 2a^2b^2 + b^4\)

Solution:

\(a^4 + 2a^2b^2 + b^4 = (a^2)^2 + 2 (a^2) (b^2) + (b^2)^2\)

\(= (a^2 + b^2)^2 \ \ \ \ \ \ \ [\because(a + b)^2 = a^2 + 2ab + b^2]\)

Question 2: Factorise

(i) \(4p^2 − 9q^2\)

Solution:

\(4p^2 − 9q^2 = (2p)^2 − (3q)^2\)

\(= (2p + 3q) (2p − 3q) [a^2 − b^2 = (a − b) (a + b)]\)

(ii) \(63a^2 − 112b^2\)

Solution:

\(63a^2 − 112b^2 = 7(9a^2 − 16b^2)\)

\(= 7[(3a)^2 − (4b)^2]\)

\(= 7(3a + 4b) (3a − 4b) [a^2 − b^2 = (a − b) (a + b)]\)

(iii) \(49x^2 − 36\)

Solution:

\(49x^2 − 36 = (7x)^2 − (6)^2\)

\(= (7x − 6) (7x + 6) [a^2 − b^2 = (a − b) (a + b)]\)

(iv) \(16x^5 − 144x^3\)

Solution:

\(16x^5 − 144x^3 = 16x^3(x^2 − 9)\)

\(= 16 x^3 [(x)^2 − (3)^2]\)

\(= 16 x^3(x − 3) (x + 3) [a^2 − b^2 = (a − b) (a + b)]\)

(v) \((l + m)^2 − (l − m)^2\)

Solution:

\((l + m)^2 − (l − m)^2 = [(l + m) − (l − m)] [(l + m) + (l − m)]\)

[Using identity \(a^2 − b^2 = (a − b) (a + b)]\)

\(= (l + m − l + m) (l + m + l − m)\)

\(= 2m × 2l\)

\(= 4ml\)

\(= 4lm\)

(vi) \(9x^2y^2 − 16\)

Solution:

\(9x^2y^2 − 16 = (3xy)^2 − (4)^2\)

\(= (3xy − 4) (3xy + 4) [a^2 − b^2 = (a − b) (a + b)]\)

(vii) \((x^2 − 2xy + y^2) − z^2\)

Solution:

\((x^2 − 2xy + y^2) − z^2 = (x − y)^2 − (z)^2 [(a − b)^2 = a^2 − 2ab + b^2]\)

\(= (x − y − z) (x − y + z) [a^2 − b^2 = (a − b) (a + b)]\)

(viii) \(25a^2 − 4b^2 + 28bc − 49c^2\)

Solution:

\(25a^2 − 4b^2 + 28bc − 49c^2 = 25a^2 − (4b^2 − 28bc + 49c^2)\)

\(= (5a)^2 − [(2b)^2 − 2 × 2b × 7c + (7c)^2]\)

\(= (5a)^2 − [(2b − 7c)^2]\)

[Using identity \((a − b)^2 = a^2 − 2ab + b^2]\)

\(= [5a + (2b − 7c)] [5a − (2b − 7c)]\)

[Using identity \(a^2 − b^2 = (a − b) (a + b)]\)

\(= (5a + 2b − 7c) (5a − 2b + 7c)\)

Question 3:Factorise the expressions

(i)\( ax^2 + bx= a × x × x + b × x = x(ax + b)\)

(ii) \(7p^2 + 21q^2= 7 × p × p + 3 × 7 × q × q = 7(p^2 + 3q^2)\)

(iii) \(2x^3 + 2xy^2 + 2xz^2= 2x(x^2 + y^2 + z^2)\)

(iv) \(am^2 + bm^2 + bn^2 + an^2 = am^2 + bm^2 + an^2 + bn^2\)

\(= m^2(a + b) + n^2(a + b)\)

\(= (a + b) (m^2 + n^2)\)

(v) \((lm + l) + m + 1 = lm + m + l + 1\)

\(= m(l + 1) + 1(l + 1)\)

\(= (l + l) (m + 1)\)

(vi)\( y (y + z) + 9 (y + z) = (y + z) (y + 9)\)

(vii) \(5y^2 − 20y − 8z + 2yz = 5y^2 − 20y + 2yz − 8z\)

\(= 5y(y − 4) + 2z(y − 4)\)

\(= (y − 4) (5y + 2z)\)

(viii) \(10ab + 4a + 5b + 2 = 10ab + 5b + 4a + 2\)

\(= 5b(2a + 1) + 2(2a + 1)\)

\(= (2a + 1) (5b + 2)\)

(ix) \(6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6\)

\(= 3x(2y − 3) − 2(2y − 3)\)

\(= (2y − 3) (3x − 2)\)

Question 4:-Factorise

(i)\( a^4 − b^4 = (a^2)^2 − (b^2)^2\)

\(= (a^2 − b^2) (a^2 + b^2)\)

\(= (a − b) (a + b) (a^2 + b^2)\)

(ii) \(p^4 − 81 = (p^2)^2 − (9)^2\)

\(= (p^2 − 9) (p^2 + 9)\)

\(= [(p)^2 − (3)^2] (p^2 + 9)\)

\(= (p − 3) (p + 3) (p^2 + 9)\)

(iii) \(x^4 − (y + z)^4 = (x^2)^2 − [(y +z)^2]^2\)

\(= [x^2 − (y + z)^2] [x^2 + (y + z)^2]\)

\(= [x − (y + z)][ x + (y + z)] [x^2 + (y + z)^2]\)

\(= (x − y − z) (x + y + z) [x^2 + (y + z)^2]\)

(iv)\( x^4 − (x − z)^4 = (x^2)^2 − [(x − z)^2]^2\)

\(= [x^2 − (x − z)^2] [x^2 + (x − z)^2]\)

\(= [x − (x − z)] [x + (x − z)] [x^2 + (x − z)^2]\)

\(= z(2x − z) [x^2 + x^2 − 2xz + z^2]\)

\(= z(2x − z) (2x^2 − 2xz + z^2)\)

(v) \(a^4 − 2a^2b^2 + b^4 = (a^2)^2 − 2 (a^2) (b^2) + (b^2)^2\)

\(= (a^2 − b^2)^2\)

\(= [(a − b) (a + b)]^2\)

\(= (a − b)^2 (a + b)^2\)

Question 5: Factorise the following expressions

(i) \(p^2 + 6p + 8\)

\(8 = 4 × 2\) and \(4 + 2 = 6\)

\(∴ p^2 + 6p + 8 = p^2 + 2p + 4p + 8\)

\(= p(p + 2) + 4(p + 2)\)

\(= (p + 2) (p + 4)\)

(ii) \(q^2 − 10q + 21\)

\(21 = (−7) × (−3)\) and \((−7) + (−3) = − 10\)

\(∴ q^2 − 10q + 21 = q^2 − 7q − 3q + 21\)

\(= q(q − 7) − 3(q − 7)\)

\(= (q − 7) (q − 3)\)

(iii) \(p^2 + 6p − 16\)

\(16 = (−2) × 8\) and \(8 + (−2) = 6\)

\(p^2 + 6p − 16 = p^2 + 8p − 2p − 16\)

\(= p(p + 8) − 2(p + 8)\)

\(= (p + 8) (p − 2)\)