RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4

Other Exercises

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.
p²q -pr²-pq + r²
Solution:
p²q -pr²-pq + r2
= p²q -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 + x²y
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
= 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 – x²y²Solution:
x³ -y² + x – x²y²⇒ x³ + x – x²y² – 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)

Question 14.
x³ – 2x²y + 3xy² – 6y³
Solution:
x³ – 2x²y + 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= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²= a²x² + b²x² + a²y² + 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 + a²x + b²x
= abx² + b²x + a²x + ab
= bx (ax + b) + a (ax + b) {(ax + b) is common}
= (ax + b) (bx + a)

Question 19.
a²x² + (ax² + 1) x + a
Solution:
a²x² + (ax² + 1) x + a
= a²x² + ax³ + x + a
= ax³ + a²x² + 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)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4 are helpful to complete your math homework.

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