Ads Blocker Image Powered by Code Help Pro

Ads Blocker Detected!!!

We have detected that you are using extensions to block ads. Please support us by disabling these ads blocker.

Mathematics – Class 8 – Chapter 14 – Factorisation – Exercise 14.2 – NCERT Exercise Solution

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)

👍👍👍

Leave a Comment

error: