Question 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
(Hint: Expand (l+m)2 first)
(viii) a4+2a2b2+b4
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
Question 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
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)
Question 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
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)
Question 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
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
Question 5. Factorise the following expressions.
(i) p2+6p+8
(ii) q2–10q+21
(iii) p2+6p–16
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)
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