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# 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)

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