plant breeding methods in asexually or clonally propagated crops
C7 7.2
1. Factorize each of the following expressions :
Question 1.
qr-pr + qs – ps
Solution:
qr- pr + qs-ps
Arranging in suitable groups = r(q-p) +s (q-p) {(q – p) is common}
= (q-p) (r + s)
Question 2.
p2
q -pr2
-pq + r2
Solution:
p2
q -pr2
-pq + r2
= p2
q -pq-pr2
+ r2
(Arranging in group)
= pq(p- 1)-r2
(p-1) {(p – 1) is common}
= (p – 1) (pq – r2
)
Question 3.
1 + x + xy + x2
y
Solution:
1 + x + xy + x 2y
= 1 (1 + x) +xy (1 +x)
= (1 + x) (1 + xy) {(1 + x) is common}
Question 4.
ax + ay – bx – by
Solution:
ax + ay – bx – by
= a (x + y) – b (x + y) {(x + y) is coinmon}
= (x+y) (a- b)
Question 5.
xa2
+ xb2
-ya2
– yb2
Solution:
xa2
+ xb2
– ya2
– yb2
= x (a2
+ b2
) -y (a2
+ b2
) {(a2
+ b2
) is common}
= {a2
+ b2
) (x -y)
Question 6.
x2
+ xy + xz + yz
Solution:
x2
+ xy + xz + yz
= x (x + y) + z(x + y) {(x + y) is common}
= (x + y) (x + z)
Question 7.
2ax + bx + 2ay + by
Solution:
2ax + bx + 2ay + by
2. = x {2a + b) + y (2a + b) {(2a + b) is common}
= (2a + b) (x + y)
Question 8.
ab- by- ay +y2
Solution:
ab – by – ay + y2
= b(a-y)-y(a-y) {(a -y) is common}
= (a-y) (b – y)
Question 9.
axy + bcxy -az- bcz
Solution:
axy + bcxy – az – bcz
= xy (a + bc) – z (a + bc) {(a + bc) is common}
= (a + bc) (xy – z)
Question 10.
lm2
– mn2
– lm + n2
Solution:
lm2
– mn2
– lm + n2
= m (lm – n2
)- 1 (lm – n2
) {(lm – n2
) is common}
= (lm – n2
) (m – 1)
Question 11.
x3
– y2
+ x – x2
y2
Solution:
x3
-y2
+ x – x2
y2
⇒ x3
+ x – x2
y2
– y2
= x(x2
+ 1)-y2
(x2
+ 1) {(x2
+ 1) is common}
= (x2
+ 1) (x -y2
)
Question 12.
6xy + 6 – 9y – 4x
Solution:
6xy + 6 – 9y – 4x
= 6 xy – 4x – 9y + 6
= 2x (3y – 2) – 3 (3y – 2) {(3y – 2) is common}
= (3y-2) (2x – 3)
Question 13.
x2
– 2ax – 2ab + bx
Solution:
x2
– 2ax – 2ab + bx
⇒ x2
– 2ax + bx – 2ab
= x (x – 2a) + b (x – 2a) {(x – 2a) is common}
= (x – 2a) (x + b)
3. Question 14.
x3
– 2x2
y + 3xy2
– 6y3
Solution:
x3
– 2x2
y + 3xy2
– 6y3
= x2
(x – 2y) + 3y2
(x – 2y) {(x – 2y) is common}
= (x – 2y) (x2
+ 3y2
)
Question 15.
abx2
+ (ay – b) x-y
Solution:
abx2
+ (ay – b) x-y
= abx2
+ ayx – bx -y
= ax (bx + y) – 1 (bx + y) {(bx +y) is common}
= (bx + y) (ax – 1)
Question 16.
(ax + by)2
+ (bx – ay)2
Solution:
(ax + by)2
+ (bx – ay)2
= a2
x2
+ b2
y2
+ 2abxy + b2
x2
+ a2
y2
– 2abxy
= a2
x2
+ b2
y2
+ b2
x2
+ a2
y2
= a2
x2
+ b2
x2
+ a2
y2
+ by2
= x2
(a2
+ b2
) + y2
(a2
+ b2
) {(a2
+ b2
) is common}
= (a2
+ b2
) (x2
+ y2
)
Question 17.
16 (a – b)3
-24 (a- b)2
Solution:
16 (a – b)3
-24 (a- b)2
HCF of 16, 24 = 8
and HCF of (a – b)3
, (a – b)2
= (a – b)2
∴16 (a – b)3
– 24 (a – b)2
= 8 (a-b)2
{2 (a-b)- 3}
{8 (a – b)2
is common}
= 8 (a – b)2
(2a – 2b – 3)
Question 18.
ab (x2
+ 1) + x (a2
+ b2
)
Solution:
ab (x2
+ 1) + x (a2
+ b2
)
= abx2
+ ab + a2
x + b2
x
= abx2
+ b2
x + a2
x + ab
= bx (ax + b) + a (ax + b) {(ax + b) is common}
= (ax + b) (bx + a)
Question 19.
a2
x2
+ (ax2
+ 1) x + a
Solution:
4. a2
x2
+ (ax2
+ 1) x + a
= a2
x2
+ ax3
+ x + a
= ax3
+ a2
x2
+ x + a
= ax2
(x + a) + 1 (x + a) {(x + a) is common}
= (x + a) (ax2
+ 1)
Question 20.
a(a- 2b -c) + 2bc
Solution:
a(a- 2b -c) + 2bc
= a2
– 2ab -ac +2bc
= a (a – 2b) – c (a – 2b) {(a – 2b) is common}
= (a – 2b) (a – c)
Question 21.
a (a + b – c)- bc
Solution:
a (a + b – c) – bc
= a2
+ ab – ac – bc
= a (a + b) – c (a + b) {(a + b) is common}
= (a + b) (a – c)
Question 22.
x2
– 11xy – x +11y
Solution:
x2
– 11xy-x + 11y
= x2
-x – 11 xy + 11 y
= x (x – 1) – 11y (x – 1) {(x – 1) is common}
= (x- 1) (x- 11y)
Question 23.
ab – a – b + 1
Solution:
ab – a-b + 1
= a (b – 1) – 1 (b – 1) {(b – 1) is common}
= (b – 1) (a – 1)
Question 24.
x2
+ y – xy – x
Solution:
x2
+ y – xy – x
= x2
– x- xy + y
= x (x – 1) – y (x – 1) {(x – 1) is common}
= (x- 1) (x-y)