Class 8 Ncert Cbse Mathematics Guide | NCERT Solutions of Factorisation – Exercise 14.2

 


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 − (lm)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, a2b2 = (ab) (a + b)}
= (2p + 3q) (2p − 3q)
(ii) 63a2 − 112b2
= 7(9a2 − 16b2)
= 7{(3a)2 − (4b)2} {Now using the formula, a2b2 = (ab) (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, a2b2 = (ab) (a + b) we can write this as}
= 16 x3(x − 3) (x + 3)
(v) (l + m)2 − (lm)2
= {(l + m) − (lm)} {(l + m) + (lm)} Now using the formula, a2b2 = (ab) (a + b)
= (l + ml + m) (l + m + lm) After simplifying we get
= 2m x 2l
= 4ml
= 4lm
(vi) 9x2y2 − 16
= (3xy)2 − (4)2 using the formula, a2b2 = (ab) (a + b)
= (3xy − 4) (3xy + 4)
(vii) (x2 − 2xy + y2) − z2
= (xy)2 − (z)2 {we get by using (ab)2 = a2 − 2ab + b2}
= (xyz) (xy + z) {by using formula a2b2 = (ab) (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) a4b4
(ii) p4 − 81
(iii) x4 − (y + z)4
(iv) x4 − (xz)4
(v) a4 − 2a2b2 + b4
Solution:
(i) a4b4
= (a2)2 − (b2)2 {a4 can be written (a2)2}
= (a2b2) (a2 + b2) {using the formula, a2b2 = (ab) (a + b)}
= (ab) (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, a2b2 = (ab) (a + b)}
= {x2 − (y + z)2} {x2 + (y + z)2}
= {x − (y + z)}{ x + (y + z)} {x2 + (y + z)2}
= (xyz) (x + y + z) {x2 + (y + z)2}
(iv) x4 − (xz)4
= (x2)2 − {(xz)2}2 [x4 can be written (x2)2 and (x – z)4 can be written {(x – z)2}2]
= {x2 − (xz)2} {x2 + (xz)2}
= {x − (xz)} {x + (xz)} {x2 + (xz)2}
= z(2xz) {x2 + x2 − 2xz + z2}
= z(2xz) (2x2 − 2xz + z2)
(v) a4 − 2a2b2 + b4
= (a2)2 − 2 (a2) (b2) + (b2)2
= (a2 b2)2
= {(ab) (a + b)}2
= (ab)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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  1. how can i print it out.can the website owner send pdf file of the chapter factorization to hizbudheen786@gmail.com.plese reply.

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