CBSE Board Class VIII, Mathematics - Factorisation
Solutions of NCERT Math Textbook Exercise 14.2
(NCERT Math Textbook Page 223, 224)
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
(viii) a4 + 2a2b2 + b4
Solution:
(i)
a2 + 8a + 16
=
(a)2 + 2 x a x 4 + (4)2
=
(a + 4)2 {since we know (a + b)2
= a2 + 2ab + b2}
(ii)
p2 − 10p + 25
=
(p)2 − 2 x p x 5 + (5)2
=
(p − 5)2 {since we know (a – b)2
= a2 – 2ab + b2}
(iii)
25m2 + 30m + 9
=
(5m)2 + 2 x 5m x 3 + (3)2
=
(5m + 3)2 {as (a + b)2 = a2
+ 2ab + b2}
(iv)
49y2 + 84yz + 36z2
=
(7y)2 + 2 x (7y) x (6z) + (6z)2
=
(7y + 6z)2
(v)
4x2 − 8x + 4
=
(2x)2 − 2 (2x) (2) + (2)2
=
(2x − 2)2 {since we know (a – b)2
= a2 – 2ab + b2}
=
{(2) (x − 1)}2 {taking 2 common from the above
expression}
=
4(x − 1)2
(vi)
121b2 − 88bc + 16c2
=
(11b)2 − 2 (11b) (4c) + (4c)2
=
(11b − 4c)2
(vii)
(l + m)2 − 4lm
{using the formula (a + b)2 =
a2 + 2ab + b2, we get}
=
l2 + 2lm + m2 − 4lm
=
l2 − 2lm + m2
=
(l − m)2 {as
(a – b)2 = a2 – 2ab + b2}
(viii)
a4 + 2a2b2 + b4
=
(a2)2 + 2
(a2) (b2) + (b2)2 {a4 can be written (a2)2}
=
(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
Solution:
(i)
4p2 − 9q2
=
(2p)2 − (3q)2 {Now using the formula, a2 − b2 = (a − b) (a + b)}
=
(2p + 3q) (2p − 3q)
(ii)
63a2 − 112b2
=
7(9a2 − 16b2)
=
7{(3a)2 − (4b)2} {Now using the formula, a2 − b2 = (a − b) (a + b)}
=
7(3a + 4b) (3a − 4b)
(iii)
49x2 − 36
=
(7x)2 − (6)2
=
(7x − 6) (7x + 6)
(iv)
16x5 − 144x3 = 16x3(x2 − 9)
=
16 x3 {(x)2 − (3)2} {Now using the formula, a2 − b2 = (a − b) (a + b) we can write this as}
=
16 x3(x − 3) (x + 3)
(v)
(l + m)2 − (l
− m)2
=
{(l + m) − (l − m)} {(l + m) + (l − m)} Now using
the formula, a2 − b2 = (a − b) (a + b)
=
(l + m − l + m) (l + m + l − m) After
simplifying we get
=
2m x 2l
=
4ml
=
4lm
(vi)
9x2y2 − 16
=
(3xy)2 − (4)2
using the formula, a2 − b2 = (a − b) (a + b)
=
(3xy − 4) (3xy + 4)
(vii)
(x2 − 2xy + y2) − z2
=
(x − y)2 − (z)2
{we get by using (a − b)2 = a2
− 2ab + b2}
=
(x − y − z) (x − y + z) {by using formula a2 − b2 = (a − b) (a + b)}
(viii)
25a2 − 4b2 + 28bc − 49c2
=
25a2 − (4b2 − 28bc + 49c2)
=
(5a)2 − {(2b)2 − 2 × 2b x 7c + (7c)2}
=
(5a)2 − {(2b − 7c)2}
=
{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
Solution:
(i)
ax2 + bx
Taking x common we get
=
x(ax + b)
(ii)
7p2 + 21q2
=
7 x p x p + 3 x 7 x q x q {After breaking the expression and then taking 7 common
we get}
=
7(p2 + 3q2)
(iii)
2x3 + 2xy2 + 2xz2
=
2x(x2 + y2
+ z2)
(iv) am2 + bm2 + bn2 + an2
=
am2 + bm2 + an2 + bn2
=
m2(a + b) + n2(a + b) Taking (a
+ b) common we get
=
(a + b) (m2
+ n2)
(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)
5y2 − 20y − 8z + 2yz
By rearranging the terms we get
=
5y2 − 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) Taking (2a + 1) common we
get
=
(2a + 1) (5b + 2)
(ix)
do yourself by taking example from above
Q 4: Factorise -
(i) a4 − b4
(ii) p4 − 81
(iii) x4 − (y + z)4
(iv) x4 − (x − z)4
(v) a4 − 2a2b2 + b4
Solution:
(i)
a4 − b4
=
(a2)2 − (b2)2 {a4 can be written (a2)2}
=
(a2 − b2) (a2 + b2) {using the formula, a2 − b2 = (a − b) (a + b)}
=
(a − b) (a + b) (a2 + b2)
(ii)
p4 − 81 = (p2)2 − (9)2
=
(p2 − 9) (p2 + 9)
=
{(p)2 − (3)2}
(p2 + 9)
=
(p − 3) (p + 3) (p2 + 9)
(iii)
x4 − (y + z)4
=
(x2)2 − {(y +z)2}2 {Now using the formula, a2 − b2
= (a − b) (a + b)}
=
{x2 − (y + z)2} {x2
+ (y + z)2}
=
{x − (y + z)}{ x + (y + z)} {x2 + (y + z)2}
=
(x − y − z) (x + y + z) {x2 + (y + z)2}
(iv)
x4 − (x − z)4
=
(x2)2 − {(x − z)2}2 [x4 can be written (x2)2
and (x – z)4 can be written {(x – z)2}2]
=
{x2 − (x − z)2} {x2
+ (x − z)2}
=
{x − (x − z)} {x + (x − z)} {x2 + (x − z)2}
=
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
=
(a − b)2 (a
+ b)2
Q 5: Factorise the
following expressions -
(i) p2 + 6p + 8
(ii) q2 − 10q + 21
(iii) p2 + 6p − 16
Solution:
(i)
p2 + 6p + 8
LCM of 8 is 4, 2 and 4 + 2 = 6. The above
expression can be written as -
=
p2 + 2p + 4p + 8
=
p(p + 2) + 4(p +
2)
=
(p + 2) (p + 4)
(ii)
q2 − 10q + 21
LCM of 21 is 7, 3 and (−7) + (−3) = − 10.
Now rewrite the above expression as -
=
q2 − 7q − 3q + 21
=
q(q − 7) − 3(q −
7)
=
(q − 7) (q − 3)
(iii)
do yourself.
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