1. Factorise the following expressions.
(i) a2+8a+16
Solution:
Here we observe that 4 + 4 = 8 and 4 × 4 = 16
Now, we split 8a
= a2 + 4a + 4a + 16
= (a2 + 4a) + (4a + 16)
= a (a + 4) + 4(a + 4)
= (a + 4) (a + 4)
= (a + 4)2
(ii) p2–10p+25
Solution:
Here we observe that, 5 + 5 = 10 and 5 × 5 = 25
Now, we split 10 p
= p2 – 5p – 5p + 25
= (p2 – 5p) + (-5p + 25)
= p (p – 5) – 5(p – 5)
= (p – 5) (p – 5)
= (p – 5)2
= p2-2×5×p+52
= (p-5)2
(iii) 25m2+30m+9
Solution:
Here we observe that, 15 + 15 = 30 and 15 × 15 = 25 × 9 = 225
Now, we split 30m
= 25m2 + 15m + 15m + 9
= (25m2 + 15m) + (15m + 9)
= 5m (5m + 3) + 3(5m + 3)
= (5m + 3) (5m + 3)
= (5m + 3)2
(iv) 49y2+84yz+36z2
Solution:
Here we observe that, 42 + 42 = 84 and 42 × 42 = 49 × 36 = 1764
Now, we split 84yz
= 49y2 + 42yz + 42yz + 36z2
= 7y (7y + 6z) +6z (7y + 6z)
= (7y + 6z) (7y + 6z)
= (7y + 6z)2
(v) 4x2–8x+4
Solution:
We can take 4 common in given expression
Now,
= 4(x2 – 2x + 1)
We split 2x, such that 1 x 1 = 1 and 1 + 1 = 2
= 4 (x2 – x – x + 1)
= 4 [x (x – 1) -1(x – 1)]
= 4 (x – 1) (x – 1)
= 4 (x – 1)2
(vi) 121b2-88bc+16c2
Solution:
Here we observe that, 44 + 44 = 88 and 44 × 44 = 121 × 16 = 1936
Now, we split 88bc
= 121b2 – 44bc – 44bc + 16c2
= 11b (11b – 4c) – 4c (11b – 4c)
= (11b – 4c) (11b – 4c)
= (11b – 4c)2
(vii) (l+m)2-4lm (Hint: Expand (l+m)2 first)
Solution:
[ Using property (a + b)2 = a2 + 2ab + b2]
After expanding (l + m)2,
l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
Now, we split 2lm
= l2 – Im – lm + m2
= l(l – m) – m(l – m)
= (l – m) (l – m)
= (l – m)2
(viii) a4+2a2b2+b4
Solution:
We split 2a2b2
= a4 + a2b2 + a2b2 + b4
= a2(a2 + b2) + b2(a2 + b2)
= (a2 + b2) (a2 + b2)
= (a2 + b2)2
2. Factorise.
(i) 4p2–9q2
Solution:
Using identity: a2-b2 = (a + b) (a – b)
= (2p)2-(3q)2
= (2p-3q) (2p+3q)
(ii) 63a2–112b2
Solution:
We take 7 as common
= 7(9a2 –16b2)
Using identity: a2 – b2 = (a + b) (a – b)
= 7((3a)2–(4b)2)
= 7(3a+4b) (3a-4b)
(iii) 49x2–36
Solution:
Using identity: a2 – b2 = (a + b) (a – b)
= (7x)2 -62
= (7x+6) (7x–6)
(iv) 16x5–144x3
Solution:
We take 16x3 as common
= 16x3(x2–9)
Using identity: a2 – b2 = (a + b) (a – b)
= 16x3 (x2 – 32)
= 16x3(x–3) (x+3)
(v) (l+m)2-(l-m)2
Solution:
Using identity: a2 – b2 = (a + b) (a – b)
= {(l+m)-(l–m)} {(l +m) + (l–m)}
Now, simplify this equation
= (l+m–l+m) (l+m+l–m)
= (2m) (2l)
= 4 ml
(vi) 9x2y2–16
Solution:
Using identity: a2 – b2 = (a + b) (a – b)
= (3xy)2-42
= (3xy–4) (3xy+4)
(vii) (x2–2xy+y2)–z2
Solution:
Using identity: a2 – b2 = (a + b) (a – b)
We can write (x2 – 2xy + y2) as (x-y)2
Now,
= (x–y)2–z2
Again, using identity: a2 – b2 = (a + b) (a – b)
= {(x–y)–z} {(x–y)+z}
= (x–y–z) (x–y+z)
(viii) 25a2–4b2+28bc–49c2
Solution:
= 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – (2b – 7c)2
Using identity: a2 – b2 = (a + b) (a – b)
= {5a – (2b – 7c)} {5a + (2b – 7c)}
= (5a – 2b + 7c) (5a + 2b – 7c)
3. Factorise the expressions.
(i)ax2+bx
Solution:
Taking x as common:
= x(ax+b)
(ii) 7p2+21q2
Solution:
Taking 7 as common:
= 7(p2+3q2)
(iii) 2x3+2xy2+2xz2
Solution:
Taking 2x as common:
= 2x(x2+y2+z2)
(iv) am2+bm2+bn2+an2
Solution:
= (am2+bm2) + (bn2+an2)
Taking m2 and n2 as common:
= m2(a+b)+n2(a+b)
= (a+b)(m2+n2)
(v) (lm+l)+m+1
Solution:
= (lm+m) + (l+1)
Taking m as common:
= m(l+1) + (l+1)
= (m+1) (l+1)
(vi) y (y + z) + 9(y + z)
Solution:
= (y + 9) (y + z)
(vii) 5y2–20y–8z+2yz
Solution:
= (5y2–20y) – (8z+2yz)
= 5y(y–4) +2z(y–4)
= (y–4) (5y+2z)
(viii) 10ab+4a+5b+2
Solution:
= (10ab+5b) + (4a+2)
= 5b(2a+1) + 2(2a+1)
= (2a+1) (5b+2)
(ix) 6xy–4y+6–9x
Solution:
= (6xy–9x) – (4y+6)
= 3x(2y–3)–2(2y–3)
= (2y–3) (3x–2)
4.Factorise.
(i) a4–b4
Solution:
We can write it as:
= (a2)2-(b2)2
Now, using identity: [a2 – b2 = (a + b) (a – b)]
= (a2-b2) (a2+b2)
= (a – b) (a + b) (a2+b2)
(ii) p4–81
Solution:
We can write it as:
= (p2)2-(9)2
Now, using identity: [a2 – b2 = (a + b) (a – b)]
= (p2-9) (p2+9)
We can write (p2 – 9 = p2 – 32)
= (p2-32) (p2+9)
=(p-3) (p+3) (p2+9)
(iii) x4–(y+z) 4
Solution:
We can write this equation as:
= (x2)2– [(y+z)2]2
Now, using identity: [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
Solution:
We can write as:
= (x2)2-{(x-z)2}2
Now, using identity: [a2 – b2 = (a + b) (a – b)]
= (x2)2 – [(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 + (x – z)2]
= (2xz – z2) [(x2 + (x – z)2]
= (2xz – z2) [(x2 + (x2 + z2 – 2xz)]
(2xz – z2) (2x2 -2xz + z2)
(v) a4–2a2b2+b4
Solution:
Using identity: [a2 – b2 = (a + b) (a – b)]
= (a2)2-2a2b2+(b2)2
= (a2-b2)2
Now,
= [(a–b) (a+b)]2
= (a – b)2 (a + b)2
5. Factorise the following expressions.
(i) p2+6p+8
Solution:
We can observe that, 8 = 4×2 and 4+2 = 6
Now, we split 6p
= p2+2p+4p+8
Taking Common terms,
= p(p+2) + 4(p+2)
= (p+2) (p+4)
(ii) q2–10q+21
Solution:
We can observe that, 21 = -7×-3 and -7+(-3) = -10
Now, split 10q
= q2–3q-7q+21
Taking common terms:
= q(q–3)–7(q–3)
= (q–7) (q–3)
(iii) p2+6p–16
Solution:
We can observe that, -16 = -2×8 and 8+(-2) = 6
Now, split 6p
= p2–2p+8p–16
Taking common terms:
= p(p–2) + 8(p–2)
= (p+8) (p–2)
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