CBSE Class 8

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3

Other Exercises

• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Factorize each of the following algebraic expressions.

Question 1.
6x (2x – y) + 7y (2x – y)
Solution:
6x (2x – y) + 7y (2x – y)
= (2x – y) (6x + 7y)
[∵ (2x – y) is common]

Question 2.
2r (y – x) + s (x – y)
Solution:
2r (y – x) + s (x – y)
-2r (x – y) +s (x – y)
= (x – y) (-2r + s)                   [(x – y) is common]
= (x-y) (s-2r)

Question 3.
7a (2x – 3) + 2b (2x – 3)
Solution:
7a (2x – 3) + 3b (2x – 3)
= (2x – 3) (7a + 3b)               [(2x – 3) is common]

Question 4.
9a (6a – 5b) – 12a² (6a – 5b)
Solution:
9a (6a – 5b) – 12a² (6a – 5b)
HCF of 9 and 12 = 3
∴ 3a (6a – 5b) (3 – 4a)
{(6a – 5b) is common}

Question 5.
5 (x – 2y)² + 3 (x – 2y)
Solution:
5 (x – 2y)² + 3 (x – 2y)
= 5 (x – 2y) (x – 2y) + 3 (x – 2y)
= (x – 2y) {5 (x – 2y) + 3}
{(x – 2y) is common}
= (x – 2y) (5x – 10y + 3)

Question 6.
16 (2l – 3m)² – 12 (3m – 2l)
Solution:
16 (2l – 3m)² – 12 (3m-2l)
= 16 (2l – 3m) (2l – 3m) + 12 (2l – 3m)
HCF of 16, 12 = 4 4 (2l-3m) {4 (2l- 3m) + 3}
{(2l – 3m) is common}
= 4 (2l -3m) (8l- 12m+ 3)

Question 7.
3a (x – 2y) – b (x – 2y)
Solution:
3a (x – 2y) – b (x – 2y)
= (x – 2y) (3a – b)
{(x – 2y) is common}

Question 8.
a² (x + y) + b² (x + y) + c² (x + y)
Solution:
a² (x + y) + b² (x + y) + c² (x + 3’)
= (x + y) (a² + b² + c²)
{(x + y) is common}

Question 9.
(x-y)² + (x -y)
Solution:
(x – y)² + (x- y) = (x – y) (x – y) + (x – y)
= (x – y) (x – y + 1)                          {(a – y) is common}

Question 10.
6 (a + 2b) – 4 (a + 2b)²
Solution:
6 (a + 2b) – 4 (a + 2b)²
= 6 (a + 2b) – 4 (a + 2b) (a + 2b)
HCF of 6, 4 = 2
= 2 {a + 2b) {3 – 2 {a + 2b)
{2 (a + b) is common}
= 2 (a + 2b) (3-2 a- 4b)

Question 11.
a (x -y) + 2b (y – x) + c (x -y)²
Solution:
a (x -y) + 2b (y – x) + c (x -y)²
= a (x – y) – 2b (x – y) + c (x – y) {x – y)
= (x – y) {x – 2b + c (x – y)}
{(a – y) is common}
= (a – y) (a – 2b + cx – cy)

Question 12.
– 4 (a – 2y)² + 8 (a – 2y)
Solution:
– 4 (x – 2y)² + 8 (x – 2y)
= – 4 (x – 2y) (x – 2y) + 8 (x – 2y)
{- 4 (x – 2y) is common}
= – 4 (x – 2y) (x – 2y – 2)
= 4 (x – 2y) (2 – x + 2y)

Question 13.
x³ (a – 2b) + a² (a – 2b)
Solution:
x³ (a – 2b) + x² (a – 2b)
HCF of x³, x² = x^2
∴ x² (a – 2b) (x + 1)
{x² (x – 2b) is common}
= x² (x – 2b) (x + 1)

Question 14.
(2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
(2x – 3y) (a + b) + (3x – 2y) (a + b)
= (a + b) {2x – 3y + 3x – 2y}
{(x + b) is common}
= (a + b) (5x – 5y)
= 5 (a + b) (x – y)

Question 15.
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
Solution:
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
= 4 (x + y) (3a – b) – 6 (x + y) (3a – 2b)
HCF of 4, 6 = 2
= 2 (x + y) {2 (3a – b) – 3 (3a – 2b)}
= 2 (x + 3) {6a – 2b – 9a + 6b}
= 2 (x +y) {-3a + 4b}
= 2 (x + y) (4b – 3a)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Other Exercises

• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Factorize each of the following quadratic polynomials by using the method of completing the square.
Question 1.
p² + 6p + 8
Solution:
p² + 6p + 8
= p² + 2 x p x 3 + 3² – 3² + 8   (completing the square)
= (p² + 6p + 3²) – 1
= (p + 3)² – 1²
= (P + 3)² – (1)²       { ∵ a² + b²  = (a+b) (a-b)}
= (p +3+1) (p + 3 -1)
= (p+4) (p+ 2)

Question 2.
q² – 10q + 21
Solution:
q² – 10q + 21
= (q)² – 2 x q x 5 + (5)² – (5)² + 21   (completing the square)
= (q)² – 2 x q x 5 + (5)² -25+21
= (q)²-2 x q x 5 + (5)² – 25 +21
= (q)²-2 x q x 5 + (5)² – 4
= (q – 5)² – (2)     {∵ a² – b² = (a + b) (a – b)}
= (q- 5 + 2) (q-5-2)
=(q- 3) (q-7)

Question 3.
4y² + 12y + 5
Solution:
4y² +12y + 5
= (2y)² + 2 x 2y x 3 + (3)² – (3)² + 5    (completing the square)
= (2y + 3)² – 9 + 5
= (2y + 3)² – 4
= (2y + 3)²-(2)²   {∵ a² – b² = (a + b) (a – b)}
= (2y + 3 + 2) (2y + 3 – 2)
= (2y + 5) (2y+ 1)

Question 4.
p² + 6p- 16
Solution:
p² + 6p – 16
= (p)² + 2 x  p x 3 + (3)² – (3)² – 16    (completing the square)
= (p)² + 2 x p x 3 + (3)² – 9 – 16
= (p + 3)² – 25
= (p + 3)² – (5)²     {∵ a² -b² = {a + b) (a – b)}
= (p + 3 + 5)(p + 3-5)
= (p + 8) (p – 2)

Question 5.
x² + 12x + 20
Solution:
x² + 12x + 20
= (x)² + 2 x x x 6 + (6)² – (6)² + 20   (completing the square)
= (x)² + 2 x x x6 + (6)² -36 + 20
= (x + 6)² -16
= (x + 6)² – (4)²   {∵ a² – b² = (a + b) (a – b)}
= (x + 6 + 4) (x + 6 – 4)
= (x + 10) (x + 2)

Question 6.
a² – 14a – 51
Solution:
a² – 14a-51
= (a)² – 2 x x 7 + (7)² – (7)² – 51       (completing the square)
= (a)² – 2 x a x 7 + (7)² – 49 – 51
= (a – 7)² – 100
= (a – 7)² – (10)²    {∵  a² – b² = (a + b) (a – b)}
= (a – 7 + 10) (a – 7 – 10)
= (a + 3) (a – 17)

Question 7.
a² + 2a – 3
Solution:
a² + 2a – 3
= (a)² + 2 x a x 1 + (1)² – (1)2 – 3   (completing the square)
= (a)² + 2 x a x 1 + (1)² – 1 – 3
= (a + 1)² – 4
= (a + 1)² – (2)² {∵ a² – b² = (a + b) (a – b)}
= (a + 1 + 2) (a + 1 – 2)
= (a + 3) (a – 1)

Question 8.
4x² – 12x + 5
Solution:
4x² – 12x + 5
= (2x)² – 2 x 2x x 3 + (3)² – (3)² + 5  (completing the square)
= (2x)² – 2 x 2x x 3 + (3)² -9 + 5
= (2x – 3)² – 4
= (2x – 3)² – (2)²      {∵ a² – b² = (a + b) (a – b)}
= (2x – 3 + 2) (2x – 3 – 2)
= (2x – 1) (2x – 5)

Question 9.
y² – 7y + 12
Solution:

Question 10.
z²-4z-12
Solution:
z² – 4z – 12
= (z)² – 2 x z x 2 + (2)² – (2)² – 12  (completing the square)
= (z)² – 2 x z x 2 + (2)² – 4 – 12
= (z-2)²-16
= (z-2)²-(4)²   {∵ a² – b² = (a + b) (a – b)}
= (z – 2 + 4) (z – 2 – 4)
= (z + 2)(z-6)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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

• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

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 – x²y²
Solution:
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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2

Other Exercises

• RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.1
• RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2
• RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.3
• RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.4

Solve each of the following equations and also check your result in each case :
Question 1.
\(\frac { 2x + 5 }{ 3 }\) = 3x – 10
Solution:
\(\frac { 2x + 5 }{ 3 }\) = \(\frac { 3x – 10 }{ 1 }\)
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5

Question 2.
\(\frac { a – 8 }{ 3 }\) = \(\frac { a – 3 }{ 2 }\)
Solution:

Question 3.
\(\frac { 7y + 2 }{ 5 }\) = \(\frac { 6y – 5 }{ 11 }\)
Solution:

Question 4.
x – 2x + 2 – \(\frac { 16 }{ 3 }\) x + 5 = 3 – \(\frac { 7 }{ 2 }\) x.
Solution:

Question 5.
\(\frac { 1 }{ 2 }\) x + 7x – 6 = 7x + \(\frac { 1 }{ 4 }\)
Solution:

Question 6.
\(\frac { 3 }{ 4 }\) x + 4x = \(\frac { 7 }{ 8 }\) + 6x – 6
Solution:

Question 7.
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10
Solution:
\(\frac { 7 }{ 2 }\) x – \(\frac { 5 }{ 2 }\) x = \(\frac { 20 }{ 3 }\) x + 10

Question 8.
\(\frac { 6x + 1 }{ 2 }\) + 1 = \(\frac { 7x – 3 }{ 3 }\)
Solution:

Question 9.
\(\frac { 3a – 2 }{ 3 }\) + \(\frac { 2a + 3 }{ 2 }\) = a + \(\frac { 7 }{ 6 }\)
Solution:

Question 10.
x – \(\frac { x – 1 }{ 2 }\) = 1 – \(\frac { x – 2 }{ 3 }\)
Solution:

Question 11.
\(\frac { 3x }{ 4 }\) – \(\frac { x – 1 }{ 2 }\) = \(\frac { x – 2 }{ 3 }\)
Solution:

Question 12.
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)
Solution:
\(\frac { 5x }{ 3 }\) – \(\frac { x – 1 }{ 4 }\) = \(\frac { x – 3 }{ 5 }\)

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.

Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24

Hope given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.1
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.3
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.4
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.6
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.7
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.8
• RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.9

Factorize the following :

Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3)        (HCF of 3, 9 = 3)

Question 2.
5x – 15x²
Solution:
5x- 15x² = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x² = x}

Question 3.
20a¹²b² – 15a⁸b⁴
Solution:
20a¹²b² – 15a⁸b⁴
{HCF of 20, 15 = 5, a¹², a⁸ = a⁸, b², b⁴ = b²}
= 5a⁸ b²(4a⁴ – 3b²)

Question 4.
72xy – 96x⁷y⁶
Solution:
72xy – 96x⁷y⁶
HCF of 72, 96 = 24 of x⁶x⁷ = x⁶, y⁷,y⁶ = y⁶
∴ 72x⁷y⁶ – 96x⁷y⁶ = 24x⁶y⁶ (3y – 4x)

Question 5.
20X³ – 40x² + 80x
Solution:
20x³ – 40x² + 80x
HCF of 20, 40,80 = 20
HCF of x³, x², x = x
∴ 20x³ – 40x² + 80x = 20x (x² – 2x + 4)

Question 6.
2x³y² – 4x²y³ + 8xy⁴
Solution:
2x³y² – 4x²y³ + 8xy⁴
HCF of 2, 4, 8 = 2
HCF of x³, x², x = 1
and HCF of y², y³, y⁴ = y²
∴ 2x³y² – 4x²y³ + 8xy⁴
= 2xy² (x² – 2xy + 4y²)

Question 7.
10m³n² + 15m⁴n – 20m²n³
Solution:
10m³n² + 15m⁴n – 20m²n³
HCF of 10, 15, 20 = 5
HCF of m³, m⁴, m² = m²
HCF of n², n, n³ = n
10m³n² + 15m⁴n – 20m²n³
5m²n(2mn + 3m²– 4n²)

Question 8.
2a⁴b⁴ – 3a³b⁵ + 4a²b⁵
Solution:
2a⁴b⁴ – 3a³b⁵ + 4a²b⁵
HCF of 2, 3, 4= 1
HCF of a⁴, a³, a² = a²
HCF of b⁴, b⁵ b⁵ = b⁴
∴ 2a⁴b⁴ – 3a³b⁵ + 4a²b⁵ = a²b⁴
(2a² – 3ab + 4b)

Question 9.
28a² + 14a²b² – 21a⁴
Solution:
28a² + 14a²b² – 21a⁴
HCF of 28, 14,21 =7
HCF of a², a², a⁴ = a²
HCF of 1, b², 1 = 1
∴ 28a² + 14a²b²-21a⁴ = 7a²
(4 + 2b² – 3a²)

Question 10.
a⁴b – 3a²b² – 6ab³
Solution:
a⁴b – 3a²b² – 6ab³
HCF of 1,3,6 = 1
HCF of a⁴, a², a = a
HCF of b, b², b³ = b
∴ a⁴b – 3a²b² – 6ab³ = ab (a³ – 3ab – 6b²)

Question 11.
2l²mn – 3lm²n + 4lmn²
Solution:
2l²mn – 3lm²n + 4lmn²
HCF 2, 3,4 = 1,
HCF of l²,l,l = l
HCF of m, m², m = m
HCF of n, n, n² = n
∴ 2l² mn – 3lm²n + 4lmn²
= lmn (21 -3m + 4n)

Question 12.
x⁴y² – x²y⁴ – x⁴y⁴
Solution:
x⁴y² – x²y⁴ – x⁴y⁴
HCF of x⁴, x², x⁴ = x²
HCF of y², y⁴, y⁴ =y²
∴ x⁴y² – x²y⁴ – x⁴y⁴ = x²y² (x² -y² -x²y²)

Question 13.
9 x²y + 3 axy
Solution:
9 x²y + 3 axy
HCF of 9, 3 = 3
HCF of x², x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x²y + 3axy = 3xy (3x + a)

Question 14.
16m – 4m²
Solution:
16m – 4m²
HCF of 16, 4 = 4
HCF of m, m² = m
∴ 16m – 4m² = 4m (4 – m)

Question 15.
-4a² + 4ab – 4ca
Solution:
-4a² + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a², a, a = a
∴ -4a² + 4ab – 4ca = -4a (a – b + c)

Question 16.
^{x2yz + xy2z + xyz2}
Solution:
x²yz + xy²z + xyz²
HCF of x², x, x = x
HCF of y,y²,y=y
HCF of z, z,z² = z
∴ x²yz + xy²z + xyz² = xyz (x + y + z)

Question 17.
ax²y + bxy² + cxyz
Solution:
ax²y + bxy² + cxyz
HCF of x², x, x = x,
HCF of y,y²,y = y
ax²y + bxy² + cxyz = xy (ax + by + cz)

Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.2 are helpful to complete your math homework.

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