Exercise 7.4
Factorize each of the following expressions :
(1) qr – pr + qs – ps
Solution:
qr – pr + qs – ps
= (qr – pr) + (qs – ps)
= r(q – p) + s(q – p)
(q – p) as the common factor
= (r + s) (q – p)
(2) p2q – pr2 – pq + r2
Solution:
p2q – pr2 – pq + r2
= (p2q – pq) + (r2 – pr2)
= pq(p – 1) + r2(1 – p)
= pq(p – 1) – r2(p – 1)
(p – 1) as the common factor
= (pq – r2)(p – 1)
(3) 1 + x + xy + x2y
Solution:
1 + x + xy + x2y
= (1 + x) + (xy + x2y)
= (1 + x) + xy(1 + x)
(1 +x) as the common factor
= (1 + xy)(1 +x)
(4) ax + ay – bx – by
Solution:
ax + ay – bx – by
= (ax + ay) – (bx + by)
= a(x + y) – b(x + y)
(x + y) as the common factor
= (a – b)(x + y)
(5) xa2 + xb2 – ya2 – yb2
Solution:
xa2 + xb2 – ya2 – yb2
= (xa2 + xb2 ) – (ya2 + yb2 )
= x(a2 + b2) – y(a2 + b2)
(a2 + b2) as the common factor
= (x – y)(a2 + b2)
(6) x2 + xy +xz + yz
Solution:
x2 + xy +xz + yz
= (x2 + xy) + (xz + yz)
= x(x + y) + z(x + y)
Here (x + y) is the common factor
= (x + z)(x + y)
= (x +y)(x +z)
(7) 2ax + bx + 2ay + by
Solution:
2ax + bx + 2ay + by
= (2ax + bx) + (2ay + by)
= x(2a + b) + y(2a + b)
Here (2a + b) as the common factor
= (x + y)(2a + b)
(8) ab – by – ay + y2
Solution:
ab – by – ay + y2
= (ab – ay) + (y2 – by)
= a(b – y) + y(y – b)
Here (b – y) as the common factor
= a(b – y) – y(b – y)
= (a – y)(b – y)
(9) axy + bcxy – az – bcz
Solution:
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]
(10) lm2 – mn2– Lm + n2
Solution:
lm2 – mn2– lm + n2
= (lm2 – lm) + (n2 – mn2)
= lm(m – 1) + n2(1 – m)
= lm(m – 1) – n2(m – 1)
Here (m – 1) a sthe common factor
= (lm – n2)(m – 1)
(11) x3 – y2 + x – x2y2
Solution:
x3 – y2 + x – x2y2
= (x3 + x) – (x2y2 + y2)
= x(x2 + 1) – y2(x2 + 1)
Here (x2+ 1) as the common factor
= (x – y2)(x2 + 1)
(12) 6xy + 6 – 9y – 4x
Solution:
6xy + 6 – 9y – 4x
= (6xy – 4x) + (6 – 9y)
= 2x (3y -2) + 3(2 – 3y)
= 2x(3y – 2) – 3(3y – 2)
= (2x – 3)(3y -2)
(13) x2 – 2ax – 2ab + bx
Solution:
x2 – 2ax – 2ab + bx
= (x2 – 2ax) + (bx – 2ab)
= x(x – 2a) + b(x – 2a)
= (x + b)(x – 2a)
(14) x3 – 2x2y + 3xy2 – 6y3
Solution:
x3 – 2x2y + 3xy2 – 6y3
= (x3 – 2x2y) + (3xy2 – 6y3)
= x2(x – 2y) + 3y2(x – 2y)
= (x2 + 3y2)(x – 2y)
(15) abx2 + (ay – b)x – y
Solution:
abx2 + (ay – b)x – y
= (abx2 – bx) + (axy – y)
= bx (ax – 1) + y(ax – 1)
= (bx + y)(ax – 1)
(16) (ax + by)2 + (bx – ay)2
Solution:
(ax + by)2 + (bx – ay)2
(using identities: (a + b)^2 = a2 + b2 + 2ab ) and (a – b)^2 = a2 + b2 – 2ab ) )
= a2x2 + 2abxy + b2y2 + b2x2 – 2abxy + a2y2
= (a2x2 + a2y2) + (b2x2 + b2y2)
= a2 (x2 + y2) + b2(x2 + y2)
= (a 2 + b2)(x2 + y2)
(17) 16(a – b)3 – 24(a – b)2
Solution:
16(a – b)3 – 24(a – b)2
= 8(a – b)2 [2(a – b) – 3]
= 8(a – b)2(2a – 2b – 3)
(18) ab(x2 + 1) + x(a2 + b2)
Solution:
ab(x2 + 1) + x(a2 + b2)
= abx2 + ab + a2x + b2x
= (abx2 + a2x) + (b2x + ab)
= ax(bx + a) + b(bx + a)
= (ax + b)(bx + a)
(19) a2x2 + (ax2 + 1)x + 1 + a
Solution:
a2x2 + (ax2 + 1)x + 1 + a
= a2x2 + ax3 + x + a
= ax2(x + a) + (x + a)
= (ax2 + 1)(x + a)
(20) a(a – 2b – c) + 2bc
Solution:
a(a – 2b – c) + 2bc
= a2 – 2ab – ac + 2bc
= (a2 – ac) + (2bc – 2ab)
= a(a – c) + 2b(c – a)
= a(a – c) – 2b(a -c)
= (a – 2b)(a – c)
(21) a(a + b – c) – bc
Solution:
a(a + b – c) – bc
= a2 + ab – ac – bc
= (a2 – ac) + (ab – bc)
= a(a – c) + b(a – c)
= (a + b)(a – c)
(22) x2 – 11xy – x + 11y
Solution:
x2 – 11xy – x + 11y
= (x2 – x) + (11y – 11xy)
= x(x – 1) + 11y(1 – x)
= x(x – 1) – 11y(x – 1)
= (x – 11y)(x – 1)
(23) ab – a – b + 1
Solution:
ab – a – b + 1
= (ab – b) + (1 – a)
= b(a – 1) + (1 – a)
= b(a – 1) – (a – 1)
= (a – 1)(b – 1)
(24) x2 + y – xy – x
Solution:
x2 + y – xy – x
= (x2 – xy) + (y – x)
= x(x – y) + (y – x)
= x(x – y) – (x – y)
= (x – 1)(x – y)
