RD Sharma Solutions Class 8 Factorization Exercise 7.4

RD Sharma Solutions Class 8 Chapter 7 Exercise 7.4

RD Sharma Class 8 Solutions Chapter 7 Ex 7.4 PDF Free Download

RD Sharma Solutions Class 8 Chapter 7 Exercise 7.4

Factorize each of the following expressions :

Q.1) qr – pr + qs – ps

Soln.:

qr – pr + qs – ps

= (qr – pr) + (qs – ps)

= r(q – p) + s(q – p)

= (r + s) (q – p) [taking (q – p) as the common factor]

Q.2) p2q – pr2 – pq + r2

Soln.:

p2q – pr2 – pq + r2

= (p2q – pq) + (r2 – pr2)

pq(p – 1) + r2(1 – p)

pq(p – 1) – r2(p – 1) [since, (1 – p) = -(p – 1)]

= (pq – r2)(p – 1) [taking (p – 1) as the common factor]

Q.3) 1 + x + xy + x2y

Soln.:

1 + x + xy + x2y

= (1 + x) + (xy + x2y)

=(1 + x) + xy(1 + x)

= (1 + xy)(1 +x) [taking (1 +x) as the common factor]

Q.4) ax + ay – bx – by

Soln.:

ax + ay – bx – by

= (ax + ay) – (bx + by)

= a(x + y) – b(x + y)

= (a – b)(x + y) [taking (x + y) as the common factor]

Q.5) xa2 + xb2 – ya2 – yb2

Soln.:

xa2 + xb2 – ya2 – yb2

= (xa2 + xb2) – (ya2 + yb2)

= x(a2 + b2) – y(a2 + b2)

= (x – y)(a2 + b2) [taking (a2 + b2) as the common factor]

Q.6) x2 + xy +xz + yz

Soln.:

x2 + xy +xz + yz

= (x2 + xy) + (xz + yz)

= x(x + y) + z(x + y)

= (x + z)(x + y) [taking (x + y) s the common factor]

= (x +y)(x +z)

Q.7) 2ax + bx + 2ay + by

Soln.:

2ax + bx + 2ay + by

= (2ax + bx) + (2ay + by)

= x(2a + b) + y(2a + b)

= (x +y)(2a + b) [taking (2a +b) as the common factor]

Q.8) ab – by – ay + y2

Soln.:

ab – by – ay + y2

= (ab – ay) + (y2 – by)

= a(b – y) + y(y – b) [since, (y – b) = – (b – y)]

= a(b – y) – y(b – y) [taking (b – y) as the common factor]

= (a – y)(b – y)

Q.9) axy + bcxy – az – bcz

Soln.:

axy + bcxy – az – bcz

= (axy + bcxy) – (az – bcz)

= xy(a + bc) – z(a + bc)

= (xy – z)(a + bc) [taking (a + bc) as the common factor]

Q.10) Lm2 – mn2– Lm + n2

Soln.:

Lm2 – mn2– Lm + n2 = (Lm2 – Lm) + (n2 – mn2)

= Lm(m – 1) + n2(1 – m)

= Lm(m – 1) – n2(m – 1) [since, (1 – m) = -(m – 1)]

= (Lm – n2)(m – 1) [taking (m – 1) a sthe common factor]

Q.11) x3 – y2 + x – x2y2

Soln.:

x3 – y2 + x – x2y2

= (x3 + x) – (x2y2 + y2)

= x(x2 + 1) – y2(x2 + 1)

= (x – y2)(x2 + 1) [taking (x2+ 1) as the common factor]

Q.12) 6xy + 6 – 9y – 4x

Soln.:

6xy + 6 – 9y – 4x = (6xy – 4x) + (6 – 9y)

= 2x (3y -2) + 3(2 – 3y)

= 2x(3y – 2) – 3(3y – 2) [since, (2 -3y) = -(3y -2)]

= (2x – 3)(3y -2) [taking (3y -2) as the common factor]

Q.13) x2 – 2ax – 2ab + bx

Soln.:

x2 – 2ax – 2ab + bx

= (x2 – 2ax) + (bx – 2ab)

= x(x – 2a) + b(x – 2a)

= (x + b)(x – 2a) [taking (x – 2a) as the common factor]

= (x – 2a)(x + b)

Q.14) x3 – 2x2y + 3xy2 – 6y3

Soln.:

x3 – 2x2y + 3xy2 – 6y3

= (x3 – 2x2y) + (3xy2 – 6y3)

= x2(x – 2y) + 3y2(x – 2y)

= (x2 + 3y2)(x – 2y) [taking (x – 2y) as the common factor]

Q.15) abx2 + (ay – b)x – y

Soln.:

abx2 + (ay – b)x – y = abx2 + axy – bx – y

= (abx2 – bx) + (axy – y)

= bx (ax – 1) + y(ax – 1)

= (bx + y)(ax – 1) [taking (ax – 1) as the common factor]

Q.16) (ax + by)2 + (bx – ay)2

Soln.:

(ax + by)2 + (bx – ay)2 = a2x2 + 2abxy + b2y2 + b2x2 – 2abxy + a2y2

= a2x2 + b2y2 + b2x2 + a2y2

= (a2x2 + a2y2) + (b2x2 + b2y2)

= a2(x2+ y2) + b2’(x2 + y2)

= (a2+ b2)(x2 + y2) [taking (x2 +y2) as the common factor]

Q.17) 16(a – b)3 – 24(a – b)2

Soln.:

16(a – b)3 – 24(a – b)2

= 8(a – b)2 [2(a -b) -3] [taking 8(a -b)2 as the common factor]

= 8(a – b)2(2a – 2b – 3)

Q.18) ab(x2 + 1) + x(a2 + b2)

Soln.:

ab(x2 + 1) + x(a2 + b2) = abx2 + ab + a2x + b2x

= (abx2 + a2x) + (b2x + ab)

= ax(bx + a) + b(bx + a)

= (ax + b)(bx + a) [taking (bx +a) as the common factor] Q.19) a2x2 + (ax2 +1)x + 1 + a

Soln.:

a2x2 + (ax2 +1)x + 1 + a = a2x2 + ax3 + x + a

= (ax3 + a2x2) + (x + a)

= ax2(x + a) + (x + a)

= (ax2 + 1)(x + a) [taking (x +a) as the common factor]

Q.20) a(a – 2b – c) + 2bc

Soln.:

a(a – 2b – c) + 2bc = a2 – 2ab – ac + 2bc

= (a2 – ac) + (2bc – 2ab)

= a(a – c) + 2b(c – a) [since, (c -a) = -(a – c)]

= a(a – c) – 2b(a -c)

= (a -2b)(a – c) [taking (a -c) as the common factor]

Q.21) a(a + b – c) – bc

Soln.:

a(a + b – c) – bc = a2 + ab – ac – bc

= (a2 – ac) + (ab – bc)

= a(a -c) + b(a – c)

= (a +b)(a – c) [taking (a -c) as the common factor]

Q.22) x2 – 11xy -x + 11y

Soln.:

x2 – 11xy -x + 11y = (x2 – x) + (11y – 11xy)

= x(x – 1) + 11y(1 – x)

= x(x – 1) – 11y(x – 1) [since, (1 – x) = – (x – 1)]

= (x – 11y)(x – 1) [taking out the common factor]

Q.23) ab – a – b + 1

Soln.:

ab – a – b + 1 = (ab – b) + (1 – a)

= b(a – 1) + (1 – a)

= b(a – 1) – (a – 1) [since, (1-a) = -(a -1)]

= (a – 1)(b – 1) [taking out the common factor (a – 1)]

Q.24) x2 + y – xy – x

Soln.:

x2 + y – xy – x = (x2 – xy) + (y -x)

= x(x – y) + (y – x)

= x(x – y) – (x – y) [(y – x) = -(x – y)]

= (x – 1)(x – y) [taking (x – y) as the common factor]

Practise This Question

Let dxx2008+x=1p ln(xq1+xr)+C where p,q,rϵN and need not be distinct, then the value of (p+q+r) equals