NCERT Solutions For Class 8 Math Chapter – Factorisation Exercise – 14.2
Q1. Factorise the following expressions.
- A2+ 8a + 16
- P2– 10p + 25
- 25m2+ 30m + 9
- 49y2+ 84yz + 36z2
- 4x2– 8x + 4
- 121b2– 88bc + 16c2
- (l + m)2– 4lm
- A4+ 2a2b2 + b4
Solution:-
- a2+ 8a + 16
A2 + 2 x 4 x a + 42
= (a + 4)2
- P2– 10p + 25
= p2 – 2 x 5 x p + 55
= (p – 5)2
( iii ) 25m2 + 30m + 9
= (5m)2 – 2 x 5m x 3 + 32
= (5m + 3)2
( iv) 49y2 + 84yz + 36z2
= (7y)2 + 2 x 7y x 6z + (6z)2
= (7y + 6z)2
(V) 4x2 – 8x + 4
= (2x)2 – 2 x 4x + 22
= (2x -2)2
(vi) 121b2 – 88bc + 16c2
=(11b)2 – 2 x 11b x 4c + (4c)2
= (11b – 4c)2
(vii) (l + m)2 – 4lm
(l + m)2 – 4lm = l2 + m2 + 2lm – 4lm
(l – m)2
(viii ) a4 + 2a2b2 + b4
= (a2)2 + 2 x a2 x b2 + (b2)2
= (a2 + b2)2
Q2. Factorise.
- 4p2– 9q2
- 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
Solution:-
- 4p2– 9q2
= (2p)2 – (3q)2
= (2p – 3q) (2p + 3q)
- 63a2– 112b2
7(9a2 – 16b2)
7((3a)2 – (4b)2)
= 7 (3a + 4b ) (3a – 4b)
- 49x2– 36
= (7a)2 – 62
= (7a + 6) (7a – 6)
- 16x5– 144x3
= 16x3(x2 – 9)
= 16x3(x – 3) (x + 3)
- (l + m)2– (l -m)2
= {(l +m) – (l – m)}{(l + m) + (l – m)}
= 4 ml
- (9x2y2– 16)
= (3xy)2 – 42
= (3xy – 4) (3xy + 4)
- (x2– 2xy + y2) – z2
= (x – y)2 – z2
= {(x – y) – z}{(x – y) + z}
= (x – y – z) (x – y + z)
- 25a2– 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – (2b – 7c)2
= (5a + 2b – 7c) (5a – 2b – 7c)
Q3. Factorise the expressions.
- Ax2+ bx
- 7p2+ 21q2
- 2x3+ 2xy2 + 2xz2
- Am2+ bm2 + bn2 + an2
- (lm + l) + m + 1
- Y(y + z) + 9(y + z)
- 5y2– 20y – 8z + 2yz
- 10ab + 4a + 5b + 2
- 6xy – 4y +6 – 9x
Solution:-
- Ax2+ bx = x(ax + b)
- 7p2+ 21q2 = 7(p2 + 3q2)
- 2x3+ 2xy2 + 2xz2 = 2x (x2 + y2 + z2)
- Am2+ bm2 + bn2 + an2 = m2 (a + b) + n2 (a + b) = (a + b) (m2 + n2)
- (lm + l) + m + 1= lm + m + l + 1 = m(l + 1) + (l + 1) = (m + 1) (l + 1)
- Y(y + z) + 9(y + z) = (y + 9) (y + z)
- 5y2– 20y – 8z + 2yz = 5y (y – 4) + 2z (y – 4) = (y – 4) (5y + 2z)
- 10ab + 4a + 5b + 2 = 5b (2a +1) + 2(2a – 1) (2a + 1) (5b + 2)
- 6xy – 4y + 6 -9x = 6xy – x – 4y + 6 = 3x(2y – 3) -2 (2y – 3) = (2y -3) (3x – 2)
Q4. Factorise.
- A4– b4
- P4– 81
- X4– (y + z)4
- X4= (x – z)4
- A4– 2a2 b2 + b4
Solution:-
- a4 – b4
= (a2)2 – (b2)2
= (a – b) (a + b) (a2 + b2)
- P4– 81
= (p2)2 – (9)2
= (p – 3) (p + 3) (p2 + 9)
- X4– (y + z)4 = [(y + z)2]2
= {x2 – (y + z)2} {x2 + ( y + z)2}
= ( x – y – z) (x + y + z) {x2 + ( y + z)2}
- X4– (x – z)4 = (x2)2 – {(x – z)2}2
= {x2 – ( x – z)2} {x2 + (x – z)2}
= z(2x – z) (2x2 – 2x2 – 2xz + z2)
- A4– 2a2b2 + b4 = (a2)2 – 2a2b2 + (b2)2
= ((a – b) (a + b))2
Q5. Factorise the following expressions.
- P2+ 6p + 8
- q2– 10q + 21
- P2+ 6p – 16
Solution:-
- P2+ 6p + 8
= p(p+2) + 4(p + 2)
= (p + 2) (p + 4)
- q2– 10q + 21
= q(q – 3) – 7(q – 3)
= (q – 7) ( q-3)
- P2+ 6p – 16
= p(p – 2) + 8(p- 2)
= (p + 8) (p – 2)