 |
VOOZH | about |
|
VOOZH | about |
Chapter 14 of the NCERT Mathematics textbook for Class 8 focuses on Factorisation a fundamental concept in algebra. Exercise 14.2 in this chapter delves deeper into the techniques and methods for the factoring algebraic expressions. This exercise helps students apply their understanding of factorization to solve problems and gain a better grasp of the polynomial expressions.
Factorization is the process of breaking down an algebraic expression into simpler components called factors which when multiplied together yield the original expression. For instance, factorizing the expression x2−9 involves expressing it as (x−3)(x+3). Factorization is essential for simplifying expressions solving equations and understanding polynomial behavior. It is a key skill in algebra that involves finding the roots or zeros of the polynomial expressions and solving quadratic equations.
Solution:
(i) a2+8a+16
Ans:
Given: a2+8a+16
Since, 8a and 16 can be substituted by 2×4×a and 42 respectively we get,
= a2+2×4×a+42
Therefore, by using the identity : (x+y)2 = x2+2xy+y2
a2+8a+16 = (a+4)2
(ii) p2–10p+25
Ans:
Given: p2–10p+25
Since, 10p and 25 can be substituted by 2×5×p and 52 respectively we get,
= p2-2×5×p+52
Therefore, by using the identity : (x-y)2 = x2-2xy+y2
p2–10p+25 = (p-5)2
(iii)25m2+30m+9
Ans:
Given: 25m2+30m+9
Since, 25m2 , 30m and 9 can be substituted by (5m)2, 2×5m×3 and 32 respectively we get,
= (5m)2 + 2×5m×3 + 32
Therefore, by using the identity: (x+y)2 = x2+2xy+y2
25m2+30m+9 = (5m+3)2
(iv)49y2+84yz+36z2
Ans:
Given: 49y2+84yz+36z2
Since, 49y2, 84yz and 36z2 can be substituted by (7y)2, 2×7y×6z and (6z)2 respectively we get,
=(7y)2+2×7y×6z+(6z)2
Therefore, by using the identity: (x+y)2 = x2+2xy+y2
49y2+84yz+36z2 = (7y+6z)2
(v)4x2–8x+4
Ans:
Given: 4x2–8x+4
Since, 4x2, 8x and 4 can be substituted by (2x)2, 2×4x and 22 respectively we get,
= (2x)2-2×4x+22
Therefore, by using the identity: (x-y)2 = x2-2xy+y2
4x2–8x+4 = (2x-2)2
(vi)121b2-88bc+16c2
Ans:
Given: 121b2-88bc+16c2
Since, 121b2, 88bc and 16c2 can be substituted by (11b)2, 2×11b×4c and (4c)2 respectively we get,
= (11b)2-2×11b×4c+(4c)2
Therefore, by using the identity: (x-y)2 = x2-2xy+y2
121b2-88bc+16c2 = (11b-4c)2
(vii)(l+m)2-4lm (Hint: Expand (l+m)2 first)
Ans:
Given: (l+m)2-4lm
By expanding (l+m)2 using identity: (x+y)2 = x2+2xy+y2 , we get,
(l+m)2-4lm = l2+m2+2lm-4lm
(l+m)2-4lm = l2+m2-2lm = l2-2lm+m2
Therefore, by using the identity: (x-y)2 = x2-2xy+y2
(l+m)2-4lm = (l-m)2
(viii)a4+2a2b2+b4
Ans:
Given: a4+2a2b2+b4
Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,
= (a2)2+2×a2×b2+(b2)2
Therefore, by using the identity: (x+y)2 = x2+2xy+y2
a4+2a2b2+b4 = (a2+b2)2
Solution:
(i) 4p2–9q2
Ans:
Given: 4p2–9q2
Since, 4p2 and 9q2 can be substituted by (2p)2 and (3q)2 respectively we get,
= (2p)2-(3q)2
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
4p2–9q2 = (2p-3q)(2p+3q)
(ii) 63a2–112b2
Ans:
Given: 63a2–112b2
63a2–112b2 = 7(9a2 –16b2)
Since, 9a2 and 16b2 can be substituted by (3a)2 and (4b)2 respectively we get,
= 7((3a)2–(4b)2)
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
= 7(3a+4b)(3a-4b)
(iii)49x2–36
Ans:
Given: 49x2–36
Since, 49x2 and 36 can be substituted by (7x)2 and 62 respectively we get,
= (7x)2 - 62
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
49x2–36 = (7x+6)(7x–6)
(iv)16x5–144x3
Ans:
Given: 16x5 – 144x3
16x5 – 144x3 = 16x3(x2–9)
Since, 9 can be substituted by 32 respectively we get,
= 16x3(x2–32)
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
16x5–144x3 = 16x3(x–3)(x+3)
(v) (l+m)2 - (l-m)2
Ans:
Given: (l+m)2 - (l-m)2
By expanding (l+m)2 - (l-m)2 using identity: x2-y2 = (x+y)(x-y) , we get,
= {(l+m)-(l–m)}{(l +m)+(l–m)}
= (l+m–l+m)(l+m+l–m)
= (2m)(2l)
(l+m)2 - (l-m)2 = 4 ml
(vi)9x2y2–16
Ans:
Given: 9x2y2–16
Since, 9x2y2 and 16 can be substituted by (3xy)2 and 42 respectively we get,
= (3xy)2-42
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
9x2y2–16 = (3xy–4)(3xy+4)
(vii)(x2–2xy+y2)–z2
Ans:
Given: (x2–2xy+y2)–z2
By compressing x2–2xy+y2 using identity: (x-y)2 = x2-2xy+y2 , we get,
(x2–2xy+y2)–z2 = (x–y)2–z2
Therefore, by using the identity: x2-y2 = (x+y)(x-y)
= {(x–y)–z}{(x–y)+z}
(x2–2xy+y2)–z2 = (x–y–z)(x–y+z)
(viii) 25a2–4b2+28bc–49c2
Ans:
Given: 25a2–4b2+28bc–49c2
25a2–4b2+28bc–49c2 = 25a2–(4b2-28bc+49c2 )
Since, 25a2, 4b2, 28bc and 49c2 can be substituted by (5a)2, (2b)2, 2(2b)(7c) and (7c)2 respectively we get,
= (5a)2-{(2b)2-2(2b)(7c)+(7c)2}
Therefore, by using the identity: (x-y)2 = x2-2xy+y2
= (5a)2-(2b-7c)2
and by using Identity: x2-y2 = (x+y)(x-y) , we get
25a2–4b2+28bc–49c2 = (5a+2b-7c)(5a-2b+7c)
Solution:
(i) ax2+bx
Ans:
Given: ax2+bx
Taking x as common, we get
ax2+bx = x(ax+b)
(ii) 7p2+21q2
Ans:
Given: 7p2+21q2
Taking 7 as common, we get,
7p2+21q2 = 7(p2+3q2)
(iii) 2x3+2xy2+2xz2
Ans:
Given: 2x3+2xy2+2xz2
Taking 2x as common, we get,
2x3+2xy2+2xz2 = 2x(x2+y2+z2)
(iv)am2+bm2+bn2+an2
Ans:
Given: am2+bm2+bn2+an2
Taking m2 and n2 as common, we get,
= m2(a+b)+n2(a+b)
Taking (a+b) as common, we get,
am2+bm2+bn2+an2 = (a+b)(m2+n2)
(v)(lm+l)+m+1
Ans:
Given: (lm+l)+m+1
= lm+m+l+1
Taking m as common, we get,
= m(l+1)+(l+1)
(lm+l)+m+1 = (m+1)(l+1)
(vi)y(y+z)+9(y+z)
Ans:
Given: y(y+z)+9(y+z)
Taking (y+z) as common, we get,
y(y+z)+9(y+z) = (y+9)(y+z)
(vii) 5y2–20y–8z+2yz
Ans:
Given: 5y2–20y–8z+2yz
Taking 5y and 2z as common, we get,
= 5y(y–4)+2z(y–4)
Taking (y-4) as common, we get,
5y2–20y–8z+2yz = (y–4)(5y+2z)
(viii)10ab+4a+5b+2
Ans:
Given: 10ab+4a+5b+2
Taking 5b and 2 as common, we get,
= 5b(2a+1)+2(2a+1)
Taking (2a+1) as common, we get,
10ab+4a+5b+2 = (2a+1)(5b+2)
(ix) 6xy–4y+6–9x
Ans:
Given: 6xy–4y+6–9x
= 6xy–9x–4y+6
Taking 3x and 2 as common, we get,
= 3x(2y–3)–2(2y–3)
Taking (2y-3) as common, we get,
6xy–4y+6–9x = (2y–3)(3x–2)
Solution:
(i)a4–b4
Ans:
Given: a4–b4
Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,
= (a2)2-(b2)2
Therefore, by using Identity: x2-y2 = (x+y)(x-y) , we get
= (a2-b2) (a2+b2)
a4–b4 = (a – b)(a + b)(a2+b2)
(ii) p4–81
Ans:
Given: p4–81
Since, p4 and 81 can be substituted by (p2)2 and (9)2 respectively we get,
= (p2)2-(9)2
Therefore, by using Identity: x2-y2 = (x+y)(x-y) , we get
= (p2-9)(p2+9)
= (p2-32)(p2+9)
p4–81 =(p-3)(p+3)(p2+9)
(iii)x4–(y+z)4
Ans:
Given: x4–(y+z)4
Since, x4 and (y+z)4 can be substituted by (x2)2 and [(y+z)2]2 respectively we get,
= (x2)2-[(y+z)2]2
Therefore, by using the identity: x2-y2 = (x+y)(x-y) , we get
= {x2-(y+z)2}{ x2+(y+z)2}
= {(x –(y+z)(x+(y+z)}{x2+(y+z)2}
x4–(y+z)4 = (x–y–z)(x+y+z) {x2+(y+z)2}
(iv) x4–(x–z)4
Ans:
Given: x4–(x–z)4
Since, x4 and (x-z)4 can be substituted by (x2)2 and [(x-z)2]2 respectively we get,
= (x2)2-{(x-z)2}2
By using Identity: x2-y2 = (x+y)(x-y) , we get
= {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)
x4–(x–z)4 = z(2x-z)( 2x2-2xz+z2)
(v)a4–2a2b2+b4
Ans:
Given: a4–2a2b2+b4
Since, a4 and b4 can be substituted by (a2)2 and (b2)2 respectively we get,
= (a2)2-2a2b2+(b2)2
Therefore, by using the identity: (x-y)2 = x2-2xy+y2
= (a2-b2)2
And by using Identity: x2-y2 = (x+y)(x-y) , we get
a4–2a2b2+b4 = ((a–b)(a+b))2
Solution:
(i) p2+6p+8
Ans:
Given: p2+6p+8
Since, 6p and 8 can be substituted by 2p+4p and 4×2 respectively we get,
= p2+2p+4p+8
Taking p and 4 common terms, we get
= p(p+2)+4(p+2)
Again taking (p+2) as common, we get
= (p+2)(p+4)
p2+6p+8 = (p+2)(p+4)
(ii)q2–10q+21
Ans:
Given: q2–10q+21
Since, 10q and 21 can be substituted by (-3q)+(-7q) and (-3)×(-7) respectively we get,
= q2–3q-7q+21
Taking q and 7 common terms, we get
= q(q–3)–7(q–3)
Again taking (q–3) as common, we get
= (q–7)(q–3)
q2–10q+21 = (q–7)(q–3)
(iii) p2+6p–16
Ans:
Given: p2+6p+16
Since, 6p and 16 can be substituted by 8p +(-2p) and (-2)×8 respectively we get,
= p2–2p+8p–16
Taking p and 8 common terms, we get
= p(p–2)+8(p–2)
Again taking (p-2) as common, we get
= (p+8)(p–2)
p2+6p–16 = (p+8)(p–2)
The Factorisation is a crucial algebraic technique that simplifies expressions and helps in the solving polynomial equations. Exercise 14.2 provides the practical problems to apply factorisation methods reinforcing students' understanding and problem-solving skills. Mastery of the factorisation not only aids in the current algebraic tasks but also lays the groundwork for the more advanced mathematical concepts.
