# RD Sharma Solutions for Class 8 Maths Chapter - 7 Factorization

RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization is the best study material for those students who are finding difficulties in solving problems. RD Sharma Class 8 maths includes all the questions provided in the textbook that is prescribed for 8th class in accordance with the CBSE Board. Subject experts at BYJUâ€™S have formulated the solutions with the utmost care to help students secure good marks in their examination. The PDFâ€™s of this chapter are available here, students can download for free from the links given below. Chapter 7- Factorization contains nine exercises and the RD Sharma Solutions present in this page provide solutions for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Definition of factors and factorization.
• Factors of a monomial, common and greatest common factor of monomials.
• Factorization of algebraic expressions when a common monomial factor occurs in each term.
• Factorization of algebraic expressions when a binomial is a common factor.
• Factorization by grouping the terms.
• Factorization of binomial expressions expressible as the difference of two squares.
• Factorization of binomial expressions expressible as a perfect square.
• Polynomials, factorization of quadratic polynomials in one variable.
• Factorization of quadratic polynomials by using the method of completing the perfect square.

## Download the pdf of RD Sharma Solutions For Class 8 Maths Chapter 7 Factorization

### EXERCISE 7.1 PAGE NO: 7.3

Find the greatest common factor (GCF/HCF) of the following polynomials: (1-14)

1. 2x2 and 12x2

Solution:

We know that the numerical coefficients of given numerical are 2 and 12

The greatest common factor of 2 and 12 is 2

The common literals appearing in given monomial is x

The smallest power of x in two monomials is 2

The monomial of common literals with smallest power is x2

âˆ´ The greatest common factor = 2x2

2. 6x3y and 18x2y3

Solution:

We know that the numerical coefficients of given numerical are 6 and18

The greatest common factor of 6 and 18 is 6

Common literals appearing in given numerical are x and y

Smallest power of x in three monomial is 2

Smallest power of y in three monomial is 1

Monomial of common literals with smallest power is x2y

âˆ´ The greatest common factor = 6x2y

3. 7x, 21x2 and 14xy2

Solution:

We know that the numerical coefficients of given numerical are 7, 21 and 14

Greatest common factor of 7, 21 and 14 is 7

Common literals appearing in given numerical are x and y

Smallest power of x in three monomials is 1

Smallest power of y in three monomials is 0

Monomials of common literals with smallest power is x

âˆ´ The greatest common factor = 7x

4. 42x2yz and 63x3y2z3

Solution:

We know that the numerical coefficients of given numerical are 42 and 63.

Greatest common factor of 42, 63 is 21.

Common literals appearing in given numerical are x, y and z

Smallest power of x in two monomials is 2

Smallest power of y in two monomials is 1

Smallest power of z in two monomials is 1

Monomials of common literals with smallest power is x2yz

âˆ´ The greatest common factor = 21x2yz

5. 12ax2, 6a2x3 and 2a3x5

Solution:

We know that the numerical coefficients of given numerical are 12, 6 and 2

Greatest common factor of 12, 6 and 2 is 2.

Common literals appearing in given numerical are a and x

Smallest power of x in three monomials is 2

Smallest power of a in three monomials is 1

Monomials of common literals with smallest power is ax2

âˆ´ The greatest common factor = 2ax2

6. 9x2, 15x2y3, 6xy2 and 21x2y2

Solution:

We know that the numerical coefficients of given numerical are 9, 15, 16 and 21

Greatest common factor of 9, 15, 16 and 21 is 3.

Common literals appearing in given numerical are x and y

Smallest power of x in four monomials is 1

Smallest power of y in four monomials is 0

Monomials of common literals with smallest power is x

âˆ´ The greatest common factor = 3x

7. 4a2b3, -12a3b, 18a4b3

Solution:

We know that the numerical coefficients of given numerical are 4, -12 and 18.

Greatest common factor of 4, -12 and 18 is 2.

Common literals appearing in given numerical are a and b

Smallest power of a in three monomials is 2

Smallest power of b in three monomials is 1

Monomials of common literals with smallest power is a2b

âˆ´ The greatest common factor = 2a2b

8. 6x2y2, 9xy3, 3x3y2

Solution:

We know that the numerical coefficients of given numerical are 6, 9 and 3

Greatest common factor of 6, 9 and 3 is 3.

Common literals appearing in given numerical are x and y

Smallest power of x in three monomials is 1

Smallest power of y in three monomials is 2

Monomials of common literals with smallest power is xy2

âˆ´ The greatest common factor = 3xy2

9. a2b3, a3b2

Solution:

We know that the numerical coefficients of given numerical are 0

Common literals appearing in given numerical are a and b

Smallest power of a in two monomials = 2

Smallest power of b in two monomials = 2

Monomials of common literals with smallest power is a2b2

âˆ´ The greatest common factor = a2b2

10. 36a2b2c4, 54a5c2, 90a4b2c2

Solution:

We know that the numerical coefficients of given numerical are 36, 54 and 90

Greatest common factor of 36, 54 and 90 is 18.

Common literals appearing in given numerical are a, b and c

Smallest power of a in three monomials is 2

Smallest power of b in three monomials is 0

Smallest power of c in three monomials is 2

Monomials of common literals with smallest power is a2c2

âˆ´ The greatest common factor = 18a2c2

11. x3, -yx2

Solution:

We know that the numerical coefficients of given numerical are 0

Common literals appearing in given numerical are x and y

Smallest power of x in two monomials is 2

Smallest power of y in two monomials is 0

Monomials of common literals with smallest power is x2

âˆ´ The greatest common factor = x2

12. 15a3, -45a2, -150a

Solution:

We know that the numerical coefficients of given numerical are 15, -45 and 150

Greatest common factor of 15, -45 and 150 is 15.

Common literals appearing in given numerical is a

Smallest power of a in three monomials is 1

Monomials of common literals with smallest power is a

âˆ´ The greatest common factor = 15a

13. 2x3y2, 10x2y3, 14xy

Solution:

We know that the numerical coefficients of given numerical are 2, 10 and 14.

Greatest common factor of 2, 10 and 14 is 2.

Common literals appearing in given numerical are x and y

Smallest power of x in three monomials is 1

Smallest power of y in three monomials is 1

Monomials of common literals with smallest power is xy

âˆ´ The greatest common factor = 2xy

14. 14x3y5, 10x5y3, 2x2y2

Solution:

We know that the numerical coefficients of given numerical are 14, 10 and 2.

Greatest common factor of 14, 10 and 2 is 2.

Common literals appearing in given numerical are x and y

Smallest power of x in three monomials is 2

Smallest power of y in three monomials is 2

Monomials of common literals with smallest power is x2y2

âˆ´ The greatest common factor = 2x2y2

Find the greatest common factor of the terms in each of the following expressions:

15. 5a4 + 10a3 – 15a2

Solution:

The greatest common factor of the three terms is 5a2

16. 2xyz + 3x2y + 4y2

Solution:

The greatest common factor of the three terms is y

17. 3a2b2 + 4b2c2 + 12a2b2c2

Solution:

The greatest common factor of the three terms is b2.

### EXERCISE 7.2 PAGE NO: 7.5

Factorize the following:

1. 3x â€“ 9

Solution:

The greatest common factor in the given two terms is 3

3x â€“ 9

3 (x – 3)

2. 5x – 15x2

Solution:

The greatest common factor in the given two terms is 5x

5x â€“ 15x2

5x (1 â€“ 3x)

3. 20a12b2 â€“ 15a8b4

Solution:

Greatest common factor in the given two terms is 5a8b2

20a12b2 â€“ 15a8b4

5a8b2Â (4a4Â – 3b2)

4. 72x6y7 – 96x7y6
Solution:

Greatest common factor in the given two terms is 24x6y6

72x6y7 – 96x7y6

24x6y6 (3y â€“ 4x)

5. 20x3Â – 40x2Â + 80x

Solution:

Greatest common factor in the given three terms is 20x

20x3Â – 40x2Â + 80x

20x (x2Â – 2x +4)

6. 2x3y2Â – 4x2y3Â + 8xy4

Solution:

Greatest common factor in the given three terms is 2xy2

2x3y2Â – 4x2y3Â + 8xy4

2xy2Â (x2Â – 2xy + 4y2)

7. 10m3n2Â + 15m4n – 20m2n3

Solution:

Greatest common factor in the given three terms is 5mn2

10m3n2Â + 15m4n – 20m2n3

5m2n (2mn + 3m2Â – 4n2)

8. 2a4b4Â – 3a3b5Â + 4a2b5

Solution:

Greatest common factor in the given three terms is a2b4

2a4b4Â – 3a3b5Â + 4a2b5

a2b4Â (2a2Â – 3ab + 4b)

9. 28a2Â + 14a2b2Â – 21a4

Solution:

Greatest common factor in the given three terms is 7a2

28a2Â + 14a2b2Â – 21a4

7a2 (4a + 2b2Â – 3a2)

10. a4b – 3a2b2Â – 6ab3

Solution:

Greatest common factor in the given three terms is ab

a4b – 3a2b2Â – 6ab3

ab (a3Â – 3ab – 6b2)

11. 2l2mn – 3lm2n + 4lmn2

Solution:

Greatest common factor in the given three terms is lmn

2l2mn – 3lm2n + 4lmn2

lmn (2l – 3m + 4n)

12. x4y2Â – x2y4Â – x4y4

Solution:

Greatest common factor in the given three terms is x2y2

x4y2Â – x2y4Â – x4y4

x2y2Â (x2Â – y2Â – x2y2)

13. 9x2y + 3axy

Solution:

Greatest common factor in the given three terms is 3xy

9x2y + 3axy

3xy (3xÂ + a)

14. 16m – 4m2

Solution:

Greatest common factor in the given two terms is 4m

16m – 4m2

4m (4 – m)

15. -4a2 + 4ab – 4ca

Solution:

Greatest common factor in the given three terms is – 4a

-4a2 + 4ab – 4ca

-4a (a – b + c)

16. x2yz + xy2z + xyz2

Solution:

Greatest common factor in the given three terms is xyz

x2yz + xy2z + xyz2

xyz (x + y +z)

17. ax2y + bxy2 + cxyz

Solution:

Greatest common factor in the given three terms is xy

ax2y + bxy2 + cxyz

xy (ax + by + cz)

### EXERCISE 7.3 PAGE NO: 7.7

Factorize each of the following algebraic expressions:

1. 6x (2x – y) + 7y (2x – y)

Solution:

We have,

6x (2x – y) + 7y (2x – y)

By taking (2x â€“ y) as common we get,

(6x + 7y) (2x â€“ y)

2. 2r (y – x) + s (x – y)

Solution:

We have,

2r (y – x) + s (x – y)

By taking (-1) as common we get,

-2r (x â€“ y) + s (x â€“ y)

By taking (x – y) as common we get,

(x â€“ y) (-2r + s)

(x â€“ y) (s â€“ 2r)

3. 7a (2x – 3) + 3b (2x – 3)

Solution:

We have,

7a (2x – 3) + 3b (2x – 3)

By taking (2x – 3) as common we get,

(7a + 3b) (2x â€“ 3)

4. 9a (6a â€“ 5b) â€“ 12a2 (6a â€“ 5b)

Solution:

We have,

9a (6a â€“ 5b) â€“ 12a2 (6a â€“ 5b)

By taking (6a â€“ 5b) as common we get,

(9a â€“ 12a2) (6a â€“ 5b)

3a(3 â€“ 4a) (6a â€“ 5b)

5. 5 (x â€“ 2y)2 + 3 (x â€“ 2y)

Solution:

We have,

5 (x â€“ 2y)2 + 3 (x â€“ 2y)

By taking (x â€“ 2y) as common we get,

(x â€“ 2y) [5 (x â€“ 2y) + 3]

(x â€“ 2y) (5x â€“ 10y + 3)

6. 16 (2l â€“ 3m)2 – 12 (3m â€“ 2l)

Solution:

We have,

16 (2l â€“ 3m)2 – 12 (3m â€“ 2l)

By taking (-1) as common we get,

16 (2l â€“ 3m)2 + 12 (2l â€“ 3m)

By taking 4(2l â€“ 3m) as common we get,

4(2l â€“ 3m) [4 (2l â€“ 3m) + 3]

4(2l â€“ 3m) (8l â€“ 12m + 3)

7. 3a (x â€“ 2y) â€“ b (x â€“ 2y)

Solution:

We have,

3a (x â€“ 2y) â€“ b (x â€“ 2y)

By taking (x â€“ 2y) as common we get,

(3a â€“ b) (x â€“ 2y)

8. a2 (x + y) + b2 (x + y) + c2 (x + y)

Solution:

We have,

a2 (x + y) + b2 (x + y) + c2 (x + y)

By taking (x + y) as common we get,

(a2Â + b2Â + c2) (x + y)

9. (x – y)2 + (x – y)

Solution:

We have,

(x – y)2 + (x – y)

By taking (x – y) as common we get,

(x â€“ y) (x â€“ y + 1)

10. 6 (a + 2b) â€“ 4 (a + 2b)2

Solution:

We have,

6 (a + 2b) â€“ 4 (a + 2b)2

By taking (a + 2b) as common we get,

[6 â€“ 4 (a + 2b)] (a + 2b)

(6 â€“ 4a â€“ 8b) (a + 2b)

2(3 â€“ 2a â€“ 4b) (a + 2b)

11. a (x – y) + 2b (y – x) + c (x â€“ y)2

Solution:

We have,

a (x – y) + 2b (y – x) + c (x â€“ y)2

By taking (-1) as common we get,

a (x – y) – 2b (x – y) + c (x â€“ y)2

By taking (x – y) as common we get,

[a â€“ 2b + c(x – y)] (x – y)

(x – y) (a â€“ 2b + cx – cy)

12. -4 (x â€“ 2y)2 + 8 (x â€“ 2y)

Solution:

We have,

-4 (x â€“ 2y)2 + 8 (x â€“ 2y)

By taking 4(x – 2y) as common we get,

[-(x â€“ 2y) + 2] 4(x – 2y)

4(x – 2y) (-x + 2y + 2)

13. x3 (a â€“ 2b) + x2 (a â€“ 2b)

Solution:

We have,

x3 (a â€“ 2b) + x2 (a â€“ 2b)

By taking x2 (a â€“ 2b) as common we get,

(x + 1) [x2 (a â€“ 2b)]

x2 (a â€“ 2b) (x + 1)

14. (2x â€“ 3y) (a + b) + (3x â€“ 2y) (a + b)

Solution:

We have,

(2x â€“ 3y) (a + b) + (3x â€“ 2y) (a + b)

By taking (a + b) as common we get,

(a + b) [(2x â€“ 3y) + (3x â€“ 2y)]

(a + b) [2x -3y + 3x â€“ 2y]

(a + b) [5x â€“ 5y]

(a + b) 5(x – y)

15. 4(x + y) (3a – b) + 6(x + y) (2b â€“ 3a)

Solution:

We have,

4(x + y) (3a – b) + 6(x + y) (2b â€“ 3a)

By taking (x + y) as common we get,

(x + y) [4(3a – b) + 6(2b â€“ 3a)]

(x + y) [12a â€“ 4b + 12b â€“ 18a]

(x + y) [-6a + 8b]

(x + y) 2(-3a + 4b)

(x + y) 2(4b â€“ 3a)

### EXERCISE 7.4 PAGE NO: 7.12

Factorize each of the following expressions:

1. qr â€“ pr + qs – ps

Solution:

We have,

qr â€“ pr + qs – ps

By grouping similar terms we get,

qr + qs â€“ pr â€“ ps

q(r + s) â€“p (r + s)

(q – p) (r + s)

2. p2q â€“ pr2 â€“ pq + r2

Solution:

We have,

p2q â€“ pr2 â€“ pq + r2

By grouping similar terms we get,

p2q â€“ pq â€“ pr2 + r2

pq(p – 1) â€“r2(p – 1)

(p – 1) (pq â€“ r2)

3. 1 + x + xy + x2y

Solution:

We have,

1 + x + xy + x2y

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

(1 + x) (1 + xy)

4. ax + ay â€“ bx â€“ by

Solution:

We have,

ax + ay â€“ bx â€“ by

a(x + y) â€“b (x + y)

(a – b) (x + y)

5. xa2 + xb2 â€“ ya2 â€“ yb2

Solution:

We have,

xa2 + xb2 â€“ ya2 â€“ yb2

x(a2 + b2) â€“y (a2 + b2)

(x – y) (a2 + b2)

6. x2 + xy + xz + yz

Solution:

We have,

x2 + xy + xz + yz

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

(x + y) (x + z)

7. 2ax + bx + 2ay + by

Solution:

We have,

2ax + bx + 2ay + by

By grouping similar terms we get,

2ax + 2ay + bx + by

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

(2a + b) (x + y)

8. ab â€“ by â€“ ay + y2

Solution:

We have,

ab â€“ by â€“ ay + y2

By grouping similar terms we get,

Ab â€“ ay â€“ by + y2

a (b â€“ y) â€“ y (b â€“ y)

(a â€“ y) (b â€“ y)

9. axy + bcxy â€“ az â€“ bcz

Solution:

We have,

axy + bcxy â€“ az â€“ bcz

By grouping similar terms we get,

axy â€“ az + bcxy â€“ bcz

a (xy â€“ z) + bc (xy â€“ z)

(a + bc) (xy â€“ z)

10. lm2 â€“ mn2 â€“ lm + n2

Solution:

We have,

lm2 â€“ mn2 â€“ lm + n2

By grouping similar terms we get,

lm2 â€“ lm â€“ mn2 + n2

lm (m â€“ 1) â€“ n2Â (m â€“ 1)

(lm â€“ n2) (m â€“ 1)

11. x3 â€“ y2 + x â€“ x2y2

Solution:

We have,

x3 â€“ y2 + x â€“ x2y2

By grouping similar terms we get,

x + x3 â€“ y2 â€“ x2y2

x (1 + x2) – y2Â (1 + x2)

(x â€“ y2) (1 + x2)

12. 6xy + 6 â€“ 9y â€“ 4x

Solution:

We have,

6xy + 6 â€“ 9y â€“ 4x

By grouping similar terms we get,

6xy â€“ 4x â€“ 9y + 6

2x (3y â€“ 2) â€“ 3 (3y â€“ 2)

(2x â€“ 3) (3y â€“ 2)

13. x2 â€“ 2ax â€“ 2ab + bx

Solution:

We have,

x2 â€“ 2ax â€“ 2ab + bx

By grouping similar terms we get,

x2 + bx â€“ 2ax â€“ 2ab

x (x + b) â€“ 2a (x + b)

(x â€“ 2a) (x + b)

14. x3 â€“ 2x2y + 3xy2 â€“ 6y3

Solution:

We have,

x3 â€“ 2x2y + 3xy2 â€“ 6y3

By grouping similar terms we get,

x3 + 3xy2 â€“ 2x2y â€“ 6y3

x (x2Â + 3y2) â€“ 2y (x2Â + 3y2)

(x â€“ 2y) (x2Â + 3y2)

15. abx2 + (ay – b) x â€“ y

Solution:

We have,

abx2 + (ay – b) x â€“ y

abx2Â + ayx â€“ bx â€“ y

By grouping similar terms we get,

abx2 â€“ bx + ayx â€“ y

bx (ax â€“ 1) + y (ax â€“ 1)

(bx + y) (ax â€“ 1)

16. (ax + by)2 + (bx – ay)2

Solution:

We have,

(ax + by)2 + (bx – ay)2

a2x2Â + b2y2Â + 2axby + b2x2Â + a2y2Â â€“ 2axby

a2x2Â + b2y2Â + b2x2Â + a2y2

By grouping similar terms we get,

a2x2Â + a2y2Â + b2y2Â + b2x2

a2Â (x2Â + y2) + b2Â (x2Â + y2)

(a2Â + b2) (x2Â + y2)

17. 16 (a – b)3 â€“ 24 (a – b)2

Solution:

We have,

16(a – b)3 â€“ 24(a – b)2

8 (a â€“ b)2Â [2 (a â€“ b) â€“ 3]

8 (a â€“ b)2Â (2a â€“ 2b â€“ 3)

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

Solution:

We have,

ab(x2 + 1) + x(a2 + b2)

abx2Â + ab + xa2Â + xb2

By grouping similar terms we get,

abx2Â + xa2 + xb2 + ab

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

(ax + b) (bx + a)

19. a2x2 + (ax2 + 1)x + a

Solution:

We have,

a2x2 + (ax2 + 1)x + a

a2x2Â + ax3Â + x + a

ax2 (aÂ + x) + 1 (xÂ + a)

(x + a) (ax2Â + 1)

20. a (a â€“ 2b â€“ c) + 2bc

Solution:

We have,

a (a â€“ 2b â€“ c) + 2bc

a2Â â€“ 2ab â€“ ac + 2bc

a (a â€“ 2b) â€“ c (a â€“ 2b)

(a â€“ 2b) (a â€“ c)

21. a (a + b â€“ c) â€“ bc

Solution:

We have,

a (a + b â€“ c) â€“ bc

a2Â + ab â€“ ac â€“ bc

a (a + b) â€“ c (a + b)

(a + b) (a â€“ c)

22. x2 â€“ 11xy â€“ x + 11y

Solution:

We have,

x2 â€“ 11xy â€“ x + 11y

By grouping similar terms we get,

x2 â€“ x â€“ 11xy + 11y

x (x â€“ 1) â€“ 11y (x â€“ 1)

(x â€“ 11y) (x â€“ 1)

23. ab â€“ a â€“ b + 1

Solution:

We have,

ab â€“ a â€“ b + 1

a (b â€“ 1) â€“ 1 (b â€“ 1)

(a â€“ 1) (b â€“ 1)

24. x2 + y â€“ xy â€“ x

Solution:

We have,

x2 + y â€“ xy â€“ x

By grouping similar terms we get,

x2 â€“ x + y â€“ xy

x (x â€“ 1) â€“ y (x â€“ 1)

(x â€“ y) (x â€“ 1)

### EXERCISE 7.5 PAGE NO: 7.17

Factorize each of the following expressions:

1. 16x2 â€“ 25y2

Solution:

We have,

16x2 â€“ 25y2

(4x)2Â â€“ (5y)2

By using the formula (a2 â€“ b2) = (a + b) (a – b) we get,

(4x + 5y) (4x â€“ 5y)

2. 27x2 â€“ 12y2

Solution:

We have,

27x2 â€“ 12y2

By taking 3 as common we get,

3 [(3x)2Â â€“ (2y)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

3 (3x + 2y) (3x â€“ 2y)

3. 144a2 â€“ 289b2

Solution:

We have,

144a2 â€“ 289b2

(12a)2Â â€“ (17b)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(12a + 17b) (12a â€“ 17b)

4. 12m2 â€“ 27

Solution:

We have,

12m2 â€“ 27

By taking 3 as common we get,

3 (4m2Â â€“ 9)

3 [(2m)2Â â€“ 32]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

3 (2m + 3) (2m â€“ 3)

5. 125x2 â€“ 45y2

Solution:

We have,

125x2 â€“ 45y2

By taking 5 as common we get,

5 (25x2Â â€“ 9y2)

5 [(5x)2Â â€“ (3y)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

5 (5x + 3y) (5x â€“ 3y)

6. 144a2 â€“ 169b2

Solution:

We have,

144a2 â€“ 169b2

(12a)2Â â€“ (13b)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(12a + 13b) (12a â€“ 13b)

7. (2a – b)2 â€“ 16c2

Solution:

We have,

(2a – b)2 â€“ 16c2

(2a â€“ b)2Â â€“ (4c)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(2a â€“ b + 4c) (2a â€“ b â€“ 4c)

8. (x + 2y)2 â€“ 4 (2x – y)2

Solution:

We have,

(x + 2y)2 â€“ 4 (2x – y)2

(x + 2y)2Â â€“ [2 (2x â€“ y)]2

By using the formula (a2 â€“ b2)= (a + b) (a – b) we get,

[(x + 2y) + 2 (2x â€“ y)] [x + 2y â€“ 2 (2x â€“ y)]

(x + 4x + 2y â€“ 2y) (x â€“ 4x + 2y + 2y)

(5x) (4y â€“ 3x)

9. 3a5 â€“ 48a3

Solution:

We have,

3a5 â€“ 48a3

By taking 3 as common we get,

3a3Â (a2Â â€“ 16)

3a3Â (a2Â â€“ 42)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

3a3Â (a + 4) (a â€“ 4)

10. a4 â€“ 16b4

Solution:

We have,

a4 â€“ 16b4

(a2)2Â â€“ (4b2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2Â + 4b2) (a2Â â€“ 4b2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2Â + 4b2) (a + 2b) (a â€“ 2b)

11. x8 â€“ 1

Solution:

We have,

x8 â€“ 1

(x4)2â€“(1)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x4Â + 1) (x4Â â€“ 1)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x4Â + 1) (x2Â + 1) (xÂ â€“ 1) (xÂ + 1)

12. 64 â€“ (a + 1)2

Solution:

We have,

64 â€“ (a + 1)2

82Â â€“ (a + 1)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[8 + (a + 1)] [8 â€“ (a + 1)]

(a + 9) (7 â€“ a)

13. 36l2 â€“ (m + n)2

Solution:

We have,

36l2 â€“ (m + n)2

(6l)2Â â€“ (m + n)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(6l + m + n) (6l â€“ m â€“ n)

14. 25x4y4 â€“ 1

Solution:

We have,

25x4y4 â€“ 1

(5x2y2)2Â â€“ (1)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(5x2y2Â â€“ 1) (5x2y2Â + 1)

15. a4 â€“ 1/b4

Solution:

We have,

a4 â€“ 1/b4

(a2)2Â â€“ (1/b2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2Â +Â 1/b2) (a2Â â€“Â 1/b2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2Â +Â 1/b2) (a â€“Â 1/b) (aÂ + 1/b)

16. x3 â€“ 144x

Solution:

We have,

x3 â€“ 144x

x [x2Â â€“ (12)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

x (x + 12) (x â€“ 12)

17. (x â€“ 4y)2 â€“ 625

Solution:

We have,

(x â€“ 4y)2 â€“ 625

(x â€“ 4y)2Â â€“ (25)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x â€“ 4y + 25) (x â€“ 4y â€“ 25)

18. 9 (a â€“ b)2 â€“ 100 (x â€“ y)2

Solution:

We have,

9 (a â€“ b)2 â€“ 100 (x â€“ y)2

[3 (a â€“ b)]2Â â€“ [10 (x â€“ y)]2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[3 (a â€“ b) + 10 (x + y)] [3 (a â€“ b) â€“ 10 (x â€“ y)] [3a â€“ 3b + 10x â€“ 10y] [3a â€“ 3b â€“ 10x + 10y]

19. (3 + 2a)2 â€“ 25a2

Solution:

We have,

(3 + 2a)2 â€“ 25a2

(3 + 2a)2Â â€“ (5a)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(3 + 2a + 5a) (3 + 2a â€“ 5a)

(3 + 7a) (3 â€“ 3a)

(3 + 7a) 3(1 â€“ a)

20. (x + y)2 â€“ (a â€“ b)2

Solution:

We have,

(x + y)2 â€“ (a â€“ b)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[(x + y) + (a â€“ b)] [(x + y) â€“ (a â€“ b)]

(x + y + a â€“ b) (x + y â€“ a + b)

21. 1/16x2y2 â€“ 4/49y2z2

Solution:

We have,

1/16x2y2 â€“ 4/49y2z2

(1/4xy)2Â â€“ (2/7yz)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(xy/4Â +Â 2yz/7) (xy/4Â – 2yz/7)

y2Â (x/4Â +Â 2/7z) (x/4Â â€“Â 2/7z)

22. 75a3b2 â€“ 108ab4

Solution:

We have,

75a3b2 â€“ 108ab4

3ab2Â (25a2Â â€“ 36b2)

3ab2Â [(5a)2Â â€“ (6b)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

3ab2Â (5a + 6b) (5a â€“ 6b)

23. x5 â€“ 16x3

Solution:

We have,

x5 â€“ 16x3

x3Â (x2Â â€“ 16)

x3Â (x2Â â€“ 42)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

x3Â (x + 4) (x â€“ 4)

24. 50/x2 â€“ 2x2/81

Solution:

We have,

50/x2 â€“ 2x2/81

2 (25/x2 â€“Â x2/81)

2 [(5/x)2Â â€“ (x/9)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

2 (5/x+Â x/9) (5/xÂ â€“Â x/9)

25. 256x3 â€“ 81x

Solution:

We have,

256x3 â€“ 81x

x (256x4Â â€“ 81)

x [(16x2)2Â â€“ 92]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

x (4x + 3) (4x – 3) (16x2 + 9)

26. a4 â€“ (2b + c)4

Solution:

We have,

a4 â€“ (2b + c)4

(a2)2Â â€“ [(2b + c)2]2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[a2Â + (2b + c)2] [a2Â â€“ (2b + c)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[a2Â + (2b + c)2] [a + 2b + c] [a â€“ 2b â€“ c]

27. (3x + 4y)4 â€“ x4

Solution:

We have,

(3x + 4y)4 â€“ x4

[(3x + 4y)2]2Â â€“ (x2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[(3x + 4y)2Â + x2] [(3x + 4y)2Â â€“ x2] [(3x + 4y)2Â + x2] [3x + 4y + x] [3x + 4y â€“ x] [(3x + 4y)2Â + x2] [4x + 4y] [2x + 4y] [(3x + 4y)2Â + x2] 8[x + 2y] [x + y]

28. p2q2 â€“ p4q4

Solution:

We have,

p2q2 â€“ p4q4

(pq)2Â â€“ (p2q2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(pq + p2q2) (pq â€“ p2q2)

p2q2Â (1 + pq) (1 â€“ pq)

29. 3x3y â€“ 243xy3

Solution:

We have,

3x3y â€“ 243xy3

3xy (x2Â â€“ 81y2)

3xy [x2Â â€“ (9y)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(3xy) (x + 9y) (x â€“ 9y)

30. a4b4 â€“ 16c4

Solution:

We have,

a4b4 â€“ 16c4

(a2b2)2Â â€“ (4c2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2b2Â + 4c2) (a2b2Â â€“ 4c2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2b2Â + 4c2) (ab + 2c) (ab â€“ 2c)

31. x4 â€“ 625

Solution:

We have,

x4 â€“ 625

(x2)2Â â€“ (25)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x2Â + 25) (x2Â â€“ 25)

(x2Â + 25) (x2Â â€“ 52)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x2Â + 25) (x + 5) (x â€“ 5)

32. x4 â€“ 1

Solution:

We have,

x4 â€“ 1

(x2)2Â â€“ (1)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x2Â + 1) (x2Â â€“ 1)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x2Â + 1) (x + 1) (x â€“ 1)

33. 49(a â€“ b)2 â€“ 25(a + b)2

Solution:

We have,

49(a â€“ b)2 â€“ 25(a + b)2

[7 (a â€“ b)]2Â â€“ [5 (a + b)]2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[7 (a â€“ b) + 5 (a + b)] [7 (a â€“ b) â€“ 5 (a + b)]

(7a â€“ 7b + 5a + 5b) (7a â€“ 7b â€“ 5a â€“ 5b)

(12a â€“ 2b) (2a â€“ 12b)

2 (6a â€“ b) 2 (a â€“ 6b)

4 (6a â€“ b) (a â€“ 6b)

34. x â€“ y – x2 + y2

Solution:

We have,

x â€“ y – x2 + y2

x â€“ y â€“ (x2Â â€“ y2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

x â€“ y â€“ (x + y) (x â€“ y)

(x â€“ y) (1 â€“ x â€“ y)

35. 16(2x â€“ 1)2 â€“ 25y2

Solution:

We have,

16(2x â€“ 1)2 â€“ 25y2

[4 (2x â€“ 1)]2Â â€“ (5y)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(8x + 5y â€“ 4) (8x â€“ 5y â€“ 4)

36. 4(xy + 1)2 â€“ 9(x â€“ 1)2

Solution:

We have,

4(xy + 1)2 â€“ 9(x â€“ 1)2

[2 (xy + 1)]2Â â€“ [3 (x â€“ 1)]2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(2xy + 2 + 3x â€“ 3) (2xy + 2 â€“ 3x + 3)

(2xy + 3x â€“ 1) (2xy â€“ 3x + 5)

37. (2x + 1)2 â€“ 9x4

Solution:

We have,

(2x + 1)2 â€“ 9x4

(2x + 1)2Â â€“ (3x2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(2x + 1 + 3x2) (2x + 1 â€“ 3x2)

(3x2Â + 2x + 1) (-3x2Â + 2x + 1)

38. x4 â€“ (2y â€“ 3z)2

Solution:

We have,

x4 â€“ (2y â€“ 3z)2

(x2)2Â â€“ (2y â€“ 3z)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(x2Â + 2y â€“ 3z) (x2Â â€“ 2y + 3z)

39. a2 â€“ b2 + a â€“ b

Solution:

We have,

a2 â€“ b2 + a â€“ b

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a + b) (a â€“ b) + (a â€“ b)

(a â€“ b) (a + b + 1)

40. 16a4 â€“ b4

Solution:

We have,

16a4 â€“ b4

(4a2)2Â â€“ (b2)2

(4a2Â + b2) (4a2Â â€“ b2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(4a2Â + b2) (2a + b) (2a â€“ b)

41. a4 â€“ 16(b â€“ c)4

Solution:

We have,

a4 â€“ 16(b â€“ c)4

(a2)2Â â€“ [4 (b â€“ c)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[a2Â + 4 (b â€“ c)2] [a2Â â€“ 4 (b â€“ c)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

[a2Â + 4 (b â€“ c)2] [(a + 2b â€“ 2c) (a â€“ 2b + 2c)]

42. 2a5 â€“ 32a

Solution:

We have,

2a5 â€“ 32a

2a (a4Â â€“ 16)

2a [(a2)2Â â€“ (4)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

2a (a2Â + 4) (a2Â â€“ 4)

2a (a2Â + 4) (a2Â â€“ 22)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

2a (a2Â + 4) (a + 2) (a â€“ 2)

43. a4b4 â€“ 81c4

Solution:

We have,

a4b4 â€“ 81c4

(a2b2)2Â â€“ (9c2)2

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2b2Â + 9c2) (a2b2Â â€“ 9c2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

(a2b2Â + 9c2) (ab + 3c) (ab â€“ 3c)

44. xy9 â€“ yx9

Solution:

We have,

xy9 â€“ yx9

-xy (x8Â â€“ y8)

-xy [(x4)2Â â€“ (y4)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

-xy (x4Â + y4) (x4Â â€“ y4)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

-xy (x4Â + y4) (x2Â + y2) (x2Â â€“ y2)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

-xy (x4Â + y4) (x2 + y2) (x + y) (x â€“ y)

45. x3 â€“ x

Solution:

We have,

x3 â€“ x

x (x2Â â€“ 1)

By using the formula (a2Â – b2)Â = (a-b) (a+b)

x (x + 1) (x â€“ 1)

46. 18a2x2 â€“ 32

Solution:

We have,

18a2x2 â€“ 32

2 [(3ax)2Â â€“ (4)2]

By using the formula (a2Â – b2)Â = (a-b) (a+b)

2 (3ax + 4) (3ax â€“ 4)

### EXERCISE 7.6 PAGE NO: 7.22

Factorize each of the following algebraic expressions:

1. 4x2Â + 12xy + 9y2

Solution:

We have,

4x2Â + 12xy + 9y2

By using the formula (x + y)2Â = x2Â + y2Â + 2xy

(2x)2Â + (3y)2Â + 2 (2x) (3y)

(2x + 3y)2

(2x + 3y) (2x + 3y)

2. 9a2Â â€“ 24ab + 16b2

Solution:

We have,

9a2Â â€“ 24ab + 16b2

By using the formula (x – y)2Â = x2Â + y2Â – 2xy

Here x = 3a, y = 4b So,

(3a)2Â + (4b)2Â â€“ 2 (3a) (4b)

(3a â€“ 4b)2

(3a â€“ 4b) (3a â€“ 4b)

3. p2q2Â â€“ 6pqr + 9r2

Solution:

We have,

p2q2Â â€“ 6pqr + 9r2

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(pq)2Â + (3r)2Â â€“ 2 (pq) (3r)

(pq â€“ 3r)2

(pq â€“ 3r) (pq â€“ 3r)

4. 36a2Â + 36a + 9

Solution:

We have,

36a2Â + 36a + 9

(6a)2 + 2 Ã— 6a Ã— 3 + 32

(6a + 3)2

5. a2Â + 2ab + b2 – 16

Solution:

We have,

a2Â + 2ab + b2 â€“ 16

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(a + b)2Â – 42

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a + b + 4) (a + b â€“ 4)

6. 9z2Â â€“ x2 + 4xy â€“ 4y2

Solution:

We have,

9z2Â â€“ x2 + 4xy â€“ 4y2

(3z)2Â â€“ [x2Â â€“ 2 (x) (2y) + (2y)2]

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(3z)2Â â€“ (x â€“ 2y)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

[(x â€“ 2y) + 3z] [â€“x + 2y + 3z)]

7. 9a4Â â€“ 24a2b2 + 16b4 â€“ 256

Solution:

We have,

9a4Â â€“ 24a2b2 + 16b4 â€“ 256

(3a2)2Â â€“ 2 (4a2) (3b2) + (4b2)2Â â€“ (16)2

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(3a2Â â€“ 4b2)2Â â€“ (16)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(3a2Â â€“ 4b2Â + 16) (3a2Â â€“ 4b2Â â€“ 16)

8. 16 â€“ a6Â + 4a3b3 – 4b6

Solution:

We have,

16 â€“ a6Â + 4a3b3 â€“ 4b6

42 â€“ [(a3)2Â â€“ 2 (a3) (2b3) + (2b3)2]

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

42Â â€“ (a3Â â€“ 2b3)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

[4 + (a3Â â€“ 2b3)] [4 â€“ (a3Â â€“ 2b3)]

9. a2Â – 2ab + b2 â€“ c2

Solution:

We have,

a2Â – 2ab + b2 â€“ c2

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(a â€“ b)2Â â€“ c2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a â€“ b + c) (a â€“ b â€“ c)

10. x2Â + 2x + 1 â€“ 9y2

Solution:

We have,

x2Â + 2x + 1 â€“ 9y2

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(x + 1)2Â â€“ (3y)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(x + 3y + 1) (x â€“ 3y + 1)

11. a2Â + 4ab + 3b2

Solution:

We have,

a2Â + 4ab + 3b2

By using factors for 3 i.e., 3 and 1

a2Â + ab + 3ab + 3b2

By grouping we get,

a (a + b) + 3b (a + b)

(a + 3b) (a + b)

12. 96 â€“ 4x – x2

Solution:

We have,

96 â€“ 4x – x2

-x2Â â€“ 4x + 96

By using factors for 96 i.e., 12 and 8

-x2Â â€“ 12x + 8x + 96

By grouping we get,

-x (x + 12) + 8 (x + 12)

(x + 12) (-x + 8)

13. a4 + 3a2 + 4

Solution:

We have,

a4 + 3a2 + 4

(a2)2Â + (a2)2Â + 2 (2a2) + 4 â€“ a2

(a2Â + 2)2Â + (-a2)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a2Â + 2 + a) (a2Â + 2 â€“ a)

(a2Â + a + 2) (a2Â â€“ a + 2)

14. 4x4 + 1

Solution:

We have,

4x4 + 1

(2x2)2Â + 1 + 4x2Â â€“ 4x2

(2x2Â + 1)2Â â€“ 4x2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(2x2Â + 1 + 2x) (2x2Â + 1 â€“ 2x)

(2x2Â + 2x + 1) (2x2Â â€“ 2x + 1)

15. 4x4 + y4

Solution:

We have,

4x4 + y4

(2x2)2Â + (y2)2Â + 4x2y2Â â€“ 4x2y2

(2x2Â + y2)2Â â€“ 4x2y2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(2x2Â + y2Â + 2xy) (2x2Â + y2Â â€“ 2xy)

16. (x + 2)4 â€“ 6(x + 2) + 9

Solution:

We have,

(x + 2)4 â€“ 6(x + 2) + 9

(x2Â + 22)2 â€“ 6x â€“ 12 + 9

(x2Â + 22 + 2(2)(x)) â€“ 6x â€“ 12 + 9

x2Â + 4 + 4x â€“ 6x â€“ 12 + 9

x2Â â€“ 2x + 1

By using the formulaÂ (aÂ â€“ b)2 = a2 + b2 â€“ 2ab

(x â€“ 1)2

17. 25 â€“ p2 â€“ q2 â€“ 2pq

Solution:

We have,

25 â€“ p2 â€“ q2 â€“ 2pq

25 â€“ (p2Â + q2Â + 2pq)

(5)2Â â€“ (p + q)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(5 + p + q) (5 â€“p â€“ q)

-(p + q + 5) (p + q â€“ 5)

18. x2 + 9y2 â€“ 6xy â€“ 25a2

Solution:

We have,

x2 + 9y2 â€“ 6xy â€“ 25a2

(x â€“ 3y)2Â â€“ (5a)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(x â€“ 3y + 5a) (x â€“ 3y â€“ 5a)

19. 49 – a2 + 8ab â€“ 16b2

Solution:

We have,

49 – a2 + 8ab â€“ 16b2

49 â€“ (a2Â â€“ 8ab + 16b2)

49 â€“ (a â€“ 4b)2

By using the formula (a2Â â€“ b2)Â = (a + b) (a – b)

(7 + a â€“ 4b) (7 â€“ a + 4b)

-(a â€“ 4b + 7) (a â€“ 4b â€“ 7)

20. a2 – 8ab + 16b2 â€“ 25c2

Solution:

We have,

a2 – 8ab + 16b2 â€“ 25c2

(a â€“ 4b)2– (5c)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a â€“ 4b + 5c) (a â€“ 4b â€“ 5c)

21. x2 – y2 + 6y – 9

Solution:

We have,

x2 – y2 + 6y â€“ 9

x2Â + 6y â€“ (y2Â â€“ 6y + 9)

x2Â â€“ (y â€“ 3)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(x + y â€“ 3) (x â€“ y + 3)

22. 25x2 â€“ 10x + 1 â€“ 36y2

Solution:

We have,

25x2 â€“ 10x + 1 â€“ 36y2

(5x)2Â â€“ 2 (5x) + 1 â€“ (6y)2

(5x â€“ 1)2Â â€“ (6y)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(5x â€“ 6y â€“ 1) (5x + 6y â€“ 1)

23. a2 â€“ b2 + 2bc â€“ c2

Solution:

We have,

a2 â€“ b2 + 2bc â€“ c2

a2Â â€“ (b2Â â€“ 2bc + c2)

a2Â â€“ (b â€“ c)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a + b â€“ c) (a â€“ b + c)

24. a2 + 2ab + b2 â€“ c2

Solution:

We have,

a2 + 2ab + b2 â€“ c2

(a + b)2Â â€“ c2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a + b + c) (a + b â€“ c)

25. 49 â€“ x2 â€“ y2 + 2xy

Solution:

We have,

49 â€“ x2 â€“ y2 + 2xy

49 â€“ (x2Â + y2Â â€“ 2xy)

72Â â€“ (x â€“ y)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

[7 + (x â€“ y)] [7 â€“ x + y]

(x â€“ y + 7) (y â€“ x + 7)

26. a2 + 4b2 â€“ 4ab â€“ 4c2

Solution:

We have,

a2 + 4b2 â€“ 4ab â€“ 4c2

a2Â â€“ 2 (a) (2b) + (2b)2Â â€“ (2c)2

(a â€“ 2b)2Â â€“ (2c)2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(a â€“ 2b + 2c) (a â€“ 2b â€“ 2c)

27. x2 â€“ y2 â€“ 4xz + 4z2

Solution:

We have,

x2 â€“ y2 â€“ 4xz + 4z2

x2Â â€“ 2 (x) (2z) + (2z)2Â â€“ y2

As (a-b)2Â = a2Â + b2Â â€“ 2ab

(x â€“ 2z)2Â â€“ y2

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

(x + y â€“ 2z) (x â€“ y â€“ 2z)

### EXERCISE 7.7 PAGE NO: 7.27

Factorize each of the following algebraic expressions:

1. x2 + 12x â€“ 45

Solution:

We have,

x2 + 12x â€“ 45

To factorize the given expression we have to find two numbers p and q such that p+q = 12 and pq = -45

So we can replace 12x by 15x â€“ 3x

-45 by 15 Ã— 3

x2Â + 12x â€“ 45 = x2Â + 15x â€“ 3x â€“ 45

= x (x + 15) â€“ 3 (x + 15)

= (x â€“ 3) (x + 15)

2. 40 + 3x â€“ x2

Solution:

We have,

40 + 3x â€“ x2

-(x2 â€“ 3x â€“ 40)

By considering, p+q = -3 and pq = -40

So we can replace -3x by 5x â€“ 8x

-40 by 5 Ã— -8

-(x2Â â€“ 3x â€“ 40) = x2Â + 5x â€“ 8x â€“ 40

= -x (x + 5) â€“ 8 (x + 5)

= -(x – 8) (x + 5)

= (-x + 8) (x + 5)

3. a2 + 3a â€“ 88

Solution:

We have,

a2 + 3a â€“ 88

By considering, p+q = 3 and pq = -88

So we can replace 3a by 11a â€“ 8a

-40 by -11 Ã— 8

a2Â + 3a â€“ 88 = a2Â + 11a â€“ 8a â€“ 88

= a (a + 11) â€“ 8 (a + 11)

= (a â€“ 8) (a + 11)

4. a2 â€“ 14a â€“ 51

Solution:

We have,

a2 â€“ 14a â€“ 51

By considering, p+q = -14 and pq = -51

So we can replace -14a by 3a â€“ 17a

-51 by -17 Ã— 3

a2Â â€“ 14a â€“ 51 = a2Â + 3a â€“ 17a â€“ 51

= a (a + 3) â€“ 17 (a + 3)

= (a â€“ 17) (a + 3)

5. x2 + 14x + 45

Solution:

We have,

x2 + 14x + 45

By considering, p+q = 14 and pq = 45

So we can replace 14x by 5x + 9x

45 by 5 Ã— 9

x2Â + 14x + 45 = x2Â + 5x + 9x + 45

= x (x + 5) â€“ 9 (x + 5)

= (x + 9) (x + 5)

6. x2 â€“ 22x + 120

Solution:

We have,

x2 â€“ 22x + 120

By considering, p+q = -22 and pq = 120

So we can replace -22x by -12x -10x

120 by -12 Ã— -10

x2Â – 22x + 120 = x2Â – 12x â€“ 10x + 120

= x (x – 12) â€“ 10 (x – 12)

= (x â€“ 10) (x – 12)

7. x2 â€“ 11x â€“ 42

Solution:

We have,

x2 â€“ 11x â€“ 42

By considering, p+q = -11 and pq = -42

So we can replace -11x by 3x -14x

-42 by 3 Ã— -14

x2Â – 11x â€“ 42 = x2Â + 3x â€“ 14x â€“ 42

= x (x + 3) â€“ 14 (x + 3)

= (x â€“ 14) (x + 3)

8. a2 + 2a â€“ 3

Solution:

We have,

a2 + 2a â€“ 3

By considering, p+q = 2 and pq = -3

So we can replace 2a by 3a -a

-3 by 3 Ã— -1

a2Â + 2a â€“ 3 = a2Â + 3a â€“ a â€“ 3

= a (a + 3) â€“ 1 (a + 3)

= (a â€“ 1) (a + 3)

9. a2 + 14a + 48

Solution:

We have,

a2 + 14a + 48

By considering, p+q = 14 and pq = 48

So we can replace 14a by 8a + 6a

48 by 8 Ã— 6

a2Â + 14a + 48 = a2Â + 8a + 6a + 48

= a (a + 8) + 6 (a + 8)

= (a + 6) (a + 8)

10. x2 â€“ 4x â€“ 21

Solution:

We have,

x2 â€“ 4x â€“ 21

By considering, p+q = -4 and pq = -21

So we can replace -4x by 3x â€“ 7x

-21 by 3 Ã— -7

x2Â + 4x â€“ 21 = x2Â + 3x â€“ 7x â€“ 21

= x (x + 3) â€“ 7 (x + 3)

= (x â€“ 7) (x + 3)

11. y2 + 5y â€“ 36

Solution:

We have,

y2 + 5y â€“ 36

By considering, p+q = 5 and pq = -36

So we can replace 5y by 9y â€“ 4y

-36 by 9 Ã— -4

y2Â + 5y â€“ 36 = y2Â + 9y â€“ 4y â€“ 36

= y (y + 9) â€“ 4 (y + 9)

= (y â€“ 4) (y + 9)

12. (a2 – 5a)2 – 36

Solution:

We have,

(a2 – 5a)2 â€“ 36

(a2 – 5a)2 – 62

By using the formula (a2Â â€“ b2) = (a+b) (a-b)

(a2Â â€“ 5a)2Â â€“ 62Â = (a2Â â€“ 5a + 6) (a2Â â€“ 5a â€“ 6)

So now we shall factorize the expression (a2Â â€“ 5a + 6)

By considering, p+q = -5 and pq = 6

So we can replace -5a by a -6a

6 by 1 Ã— -6

a2Â -5a â€“ 6 = a2Â + a â€“ 6a â€“ 6

= a (a + 1) -6(a + 1)

= (a â€“ 6) (a + 1)

So now we shall factorize the expression (a2Â â€“ 5a + 6)

By considering, p+q = -5 and pq = -6

So we can replace -5a by -2a -3a

6 by -2 Ã— -3

a2Â -5a + 6 = a2Â â€“ 2a â€“ 3a + 6

= a (a â€“ 2) -3 (a – 2)

= (a â€“ 3) (a â€“ 2)

âˆ´ (a2 – 5a)2 â€“ 36 = (a2Â â€“ 5a + 6) (a2Â â€“ 5a â€“ 6)

= (a + 1) (a â€“ 6) (a â€“ 2) (a â€“ 3)

13. (a + 7) (a â€“ 10) + 16

Solution:

We have,

(a + 7) (a â€“ 10) + 16

a2 â€“ 10a + 7a â€“ 70 + 16

a2 â€“ 3a â€“ 54

By considering, p+q = -3 and pq = -54

So we can replace -3a by 6a â€“ 9a

-54 by 6 Ã— -9

a2 â€“ 3a â€“ 54 = a2Â + 6a â€“ 9a â€“ 54

= a (a + 6) -9 (a + 6)

= (a – 9) (a + 6)

### EXERCISE 7.8 PAGE NO: 7.30

Resolve each of the following quadratic trinomials into factors:

1. 2x2Â + 5x + 3

Solution:

We have,

2x2Â + 5x + 3

The coefficient of x2 is 2

The coefficient of x is 5

Constant term is 3

We shall split up the center term i.e., 5 into two parts such that their sum p+q is 5 and product pq = 2 Ã— 3 is 6

So, we express the middle term 5x as 2x + 3x

2x2Â + 5x + 3 = 2x2Â + 2x + 3x + 3

= 2x (x + 1) + 3 (x + 1)

= (2x + 3) (x + 1)

2. 2x2Â – 3x – 2

Solution:

We have,

2x2Â – 3x â€“ 2

The coefficient of x2 is 2

The coefficient of x is -3

Constant term is -2

So, we express the middle term -3x as -4x + x

2x2Â – 3x â€“ 2 = 2x2Â – 4x + x â€“ 2

= 2x (x â€“ 2) + 1 (x â€“ 2)

= (x â€“ 2) (2x + 1)

3. 3x2Â + 10x + 3

Solution:

We have,

3x2Â + 10x + 3

The coefficient of x2 is 3

The coefficient of x is 10

Constant term is 3

So, we express the middle term 10x as 9x + x

3x2Â + 10x + 3 = 3x2Â + 9x + x + 3

= 3x (x + 3) + 1 (x + 3)

= (3x + 1) (x + 3)

4. 7x â€“ 6 â€“ 2x2

Solution:

We have,

7x â€“ 6 â€“ 2x2

â€“ 2x2 + 7x â€“ 6

2x2 – 7x + 6

The coefficient of x2 is 2

The coefficient of x is -7

Constant term is 6

So, we express the middle term -7x as -4x – 3x

2x2Â – 7x + 6 = 2x2Â – 4x – 3x + 6

= 2x (x â€“ 2) – 3 (x â€“ 2)

= (x â€“ 2) (2x â€“ 3)

5. 7x2Â – 19x – 6

Solution:

We have,

7x2Â – 19x â€“ 6

The coefficient of x2 is 7

The coefficient of x is -19

Constant term is -6

So, we express the middle term -19x as 2x â€“ 21x

7x2Â – 19x â€“ 6 = 7x2Â + 2x – 21x â€“ 6

= x (7x + 2) – 3 (7x + 2)

= (7x + 2) (x â€“ 3)

6. 28 â€“ 31x â€“ 5x2

Solution:

We have,

28 â€“ 31x â€“ 5x2

â€“ 5x2 -31x + 28

5x2 + 31x – 28

The coefficient of x2 is 5

The coefficient of x is 31

Constant term is -28

So, we express the middle term 31x as -4x + 35x

5x2Â + 31x – 28 = 5x2Â – 4x + 35x – 28

= x (5x â€“ 4) + 7 (5x â€“ 4)

= (x + 7) (5x – 4)

7. 3 + 23y â€“ 8y2

Solution:

We have,

3 + 23y â€“ 8y2

â€“ 8y2 + 23y + 3

8y2 – 23y – 3

The coefficient of y2 is 8

The coefficient of y is -23

Constant term is -3

So, we express the middle term -23y as -24y + y

8y2Â – 23y – 3 = 8y2Â – 24y + y – 3

= 8y (y â€“ 3) + 1 (y â€“ 3)

= (8y + 1) (y â€“ 3)

8. 11x2 â€“ 54x + 63

Solution:

We have,

11x2 â€“ 54x + 63

The coefficient of x2 is 11

The coefficient of x is -54

Constant term is 63

So, we express the middle term -54x as -33x â€“ 21x

11x2Â â€“ 54x + 63 = 11x2Â – 33x – 21x + 63

= 11x (x â€“ 3) – 21 (x â€“ 3)

= (11x â€“ 21) (x â€“ 3)

9. 7x â€“ 6x2 + 20

Solution:

We have,

7x â€“ 6x2 + 20

â€“ 6x2 + 7x + 20

6x2 – 7x – 20

The coefficient of x2 is 6

The coefficient of x is -7

Constant term is -20

So, we express the middle term -7x as -15x + 8x

6x2Â – 7x – 20 = 6x2Â – 15x + 8x – 20

= 3x (2x â€“ 5) + 4 (2x â€“ 5)

= (3x + 4) (2x – 5)

10. 3x2 + 22x + 35

Solution:

We have,

3x2 + 22x + 35

The coefficient of x2 is 3

The coefficient of x is 22

Constant term is 35

So, we express the middle term 22x as 15x + 7x

3x2Â + 22x + 35 = 3x2Â + 15x + 7x + 35

= 3x (x + 5) + 7 (x + 5)

= (3x + 7) (x+ 5)

11. 12x2 – 17xy + 6y2

Solution:

We have,

12x2 – 17xy + 6y2

The coefficient of x2 is 12

The coefficient of x is -17y

Constant term is 6y2

So, we express the middle term -17xy as -9xy – 8xy

12x2Â -17xy+ 6y2Â = 12x2Â – 9xy – 8xy + 6y2

= 3x (4x â€“ 3y) â€“ 2y (4x â€“ 3y)

= (3x â€“ 2y) (4x â€“ 3y)

12. 6x2 – 5xy – 6y2

Solution:

We have,

6x2 – 5xy – 6y2

The coefficient of x2 is 6

The coefficient of x is -5y

Constant term is -6y2

So, we express the middle term -5xy as 4xy – 9xy

6x2Â -5xy- 6y2Â = 6x2Â + 4xy – 9xy – 6y2

= 2x (3x + 2y) -3y (3x + 2y)

= (2x â€“ 3y) (3x + 2y)

13. 6x2 – 13xy + 2y2

Solution:

We have,

6x2 – 13xy + 2y2

The coefficient of x2 is 6

The coefficient of x is -13y

Constant term is 2y2

So, we express the middle term -13xy as -12xy â€“ xy

6x2Â -13xy+ 2y2Â = 6x2Â – 12xy – xy + 2y2

= 6x (x â€“ 2y) â€“ y (x â€“ 2y)

= (6x â€“ y) (x – 2y)

14. 14x2 + 11xy – 15y2

Solution:

We have,

14x2 + 11xy – 15y2

The coefficient of x2 is 14

The coefficient of x is 11y

Constant term is -15y2

So, we express the middle term 11xy as 21xy â€“ 10xy

14x2Â + 11xy- 15y2Â = 14x2Â + 21xy – 10xy – 15y2

= 2x (7x â€“ 5y) + 3y (7x â€“ 5y)

= (2x + 3y) (7x – 5y)

15. 6a2 + 17ab â€“ 3b2

Solution:

We have,

6a2 + 17ab â€“ 3b2

The coefficient of a2 is 6

The coefficient of a is 17b

Constant term is -3b2

So, we express the middle term 17ab as 18ab â€“ ab

6a2Â +17abâ€“ 3b2Â = 6a2Â + 18ab – ab â€“ 3b2

= 6a (a + 3b) â€“ b (a + 3b)

= (6a â€“ b) (a + 3b)

16. 36a2 + 12abc â€“ 15b2c2

Solution:

We have,

36a2 + 12abc â€“ 15b2c2

The coefficient of a2 is 36

The coefficient of a is 12bc

Constant term is -15b2c2

So, we express the middle term 12abc as 30abc â€“ 18abc

36a2Â â€“12abcâ€“ 15b2c2Â = 36a2Â + 30abc â€“ 18abc â€“ 15b2c2

= 6a (6a + 5bc) â€“ 3bc (6a + 5bc)

= (6a + 5bc) (6a â€“ 3bc)

= (6a + 5bc) 3(2a – bc)

17. 15x2 â€“ 16xyz â€“ 15y2z2

Solution:

We have,

15x2 â€“ 16xyz â€“ 15y2z2

The coefficient of x2 is 15

The coefficient of x is -16yz

Constant term is -15y2z2

So, we express the middle term -16xyz as -25xyz + 9xyz

15x2Â -16xyz- 15y2z2Â = 15x2Â – 25yz + 9yz – 15y2z2

= 5x (3x â€“ 5yz) + 3yz (3x â€“ 5yz)

= (5x + 3yz) (3x – 5yz)

18. (x â€“ 2y)2 â€“ 5 (x â€“ 2y) + 6

Solution:

We have,

(x â€“ 2y)2 â€“ 5 (x â€“ 2y) + 6

The coefficient of (x-2y)2 is 1

The coefficient of (x-2y) is -5

Constant term is 6

So, we express the middle term -5(x â€“ 2y) as -2(x â€“ 2y) -3(x â€“ 2y)

(x â€“ 2y)2Â â€“ 5 (x â€“ 2y) + 6 = (x â€“ 2y)2Â â€“ 2 (x â€“ 2y) â€“ 3 (x â€“ 2y) + 6

= (x â€“ 2y – 2) (x – 2y – 3)

19. (2a â€“ b)2 + 2 (2a â€“ b) â€“ 8

Solution:

We have,

(2a â€“ b)2 + 2 (2a â€“ b) â€“ 8

The coefficient of (2a-b)2 is 1

The coefficient of (2a-b) is 2

Constant term is -8

So, we express the middle term 2(2a â€“ b) as 4 (2a â€“b) â€“ 2 (2a â€“ b)

(2a â€“ b)2Â + 2 (2a â€“ b) â€“ 8 = (2a â€“ b)2Â + 4 (2a â€“ b) â€“ 2 (2a â€“ b) – 8

= (2a â€“ b) (2a â€“ b + 4) â€“ 2 (2a â€“ b + 4)

= (2a â€“ b + 4) (2a â€“ b â€“ 2)

### EXERCISE 7.9 PAGE NO: 7.32

Factorize each of the following quadratic polynomials by using the method of completing the square:

1. p2 + 6p + 8

Solution:

We have,

p2 + 6p + 8

Coefficient of p2Â is unity. So, we add and subtract square of half of coefficient of p.

p2Â + 6p + 8 = p2Â + 6p + 32Â â€“ 32Â + 8 (Adding and subtracting 32)

= (p + 3)2Â â€“ 12Â (By completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (p + 3 â€“ 1) (p + 3 + 1)

= (p + 2) (p + 4)

2. q2 â€“ 10q + 21

Solution:

We have,

q2 â€“ 10q + 21

Coefficient of q2Â is unity. So, we add and subtract square of half of coefficient of q.

q2Â â€“ 10q + 21 = q2Â â€“ 10q+ 52Â â€“ 52Â + 21 (Adding and subtracting 52)

= (q â€“ 5)2Â â€“ 22Â (By completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (q â€“ 5 â€“ 2) (q â€“ 5 + 2)

= (q â€“ 3) (q â€“ 7)

3. 4y2 + 12y + 5

Solution:

We have,

4y2 + 12y + 5

4(y2 + 3y + 5/4)

Coefficient of y2Â is unity. So, we add and subtract square of half of coefficient of y.

4(y2 + 3y + 5/4) = 4 [y2Â + 3y + (3/2)2Â â€“ (3/2)2Â +Â 5/4] (Adding and subtracting (3/2)2)

= 4 [(y +Â 3/2)2Â â€“ 12] (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= 4 (y +Â 3/2Â + 1) (y +Â 3/2Â â€“ 1)

= 4 (y + 1/2) (y + 5/2) (by taking LCM)

= 4 [(2y + 1)/2] [(2y + 5)/2]

= (2y + 1) (2y + 5)

4. p2 + 6p – 16

Solution:

We have,

p2 + 6p – 16

Coefficient of p2Â is unity. So, we add and subtract square of half of coefficient of p.

p2 + 6p â€“ 16 = p2Â + 6p + 32Â â€“ 32Â â€“ 16 (Adding and subtracting 32)

= (p + 3)2Â â€“ 52Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (p + 3 + 5) (p + 3 â€“ 5)

= (p + 8) (p â€“ 2)

5. x2 + 12x + 20

Solution:

We have,

x2 + 12x + 20

Coefficient of x2Â is unity. So, we add and subtract square of half of coefficient of x.

x2 + 12x + 20 = x2Â + 12x + 62Â â€“ 62Â + 20 (Adding and subtracting 62)

= (x + 6)2Â â€“ 42Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (x + 6 + 4) (x + 6 â€“ 4)

= (x + 2) (x + 10)

6. a2 â€“ 14a â€“ 51

Solution:

We have,

a2 â€“ 14a â€“ 51

Coefficient of a2Â is unity. So, we add and subtract square of half of coefficient of a.

a2 â€“ 14a â€“ 51 = a2Â â€“ 14a + 72Â â€“ 72Â â€“ 51 (Adding and subtracting 72)

= (a â€“ 7)2Â â€“ 102Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (a â€“ 7 + 10) (9 â€“ 7 â€“ 10)

= (a â€“ 17) (a + 3)

7. a2 + 2a â€“ 3

Solution:

We have,

a2 + 2a â€“ 3

Coefficient of a2Â is unity. So, we add and subtract square of half of coefficient of a.

a2 + 2a â€“ 3 = a2Â + 2a + 12Â â€“ 12Â â€“ 3 (Adding and subtracting 12)

= (a + 1)2Â â€“ 22Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (a + 1 + 2) (a + 1 â€“ 2)

= (a + 3) (a â€“ 1)

8. 4x2 – 12x + 5

Solution:

We have,

4x2 – 12x + 5

4(x2 – 3x + 5/4)

Coefficient of x2Â is unity. So, we add and subtract square of half of coefficient of x.

4(x2 â€“ 3x + 5/4) = 4 [x2Â â€“ 3x + (3/2)2Â â€“ (3/2)2Â +Â 5/4] (Adding and subtracting (3/2)2)

= 4 [(x –Â 3/2)2Â â€“ 12] (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= 4 (x –Â 3/2Â + 1) (x –Â 3/2Â â€“ 1)

= 4 (x â€“ 1/2) (x â€“ 5/2) (by taking LCM)

= 4 [(2x-1)/2] [(2x – 5)/2]

= (2x – 5) (2x – 1)

9. y2 – 7y + 12

Solution:

We have,

y2 – 7y + 12

Coefficient of y2Â is unity. So, we add and subtract square of half of coefficient of y.

y2 – 7y + 12 = y2Â â€“ 7y + (7/2)2Â â€“ (7/2)2Â + 12 [Adding and subtracting (7/2)2]

= (y â€“Â 7/2)2Â â€“ (7/2)2Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (y â€“Â (7/2-Â 1/2)) (y â€“Â (7/2Â + 1/2))

= (y â€“ 3) (y â€“ 4)

10. z2 â€“ 4z – 12

Solution:

We have,

z2 â€“ 4z â€“ 12

Coefficient of z2Â is unity. So, we add and subtract square of half of coefficient of z.

z2 â€“ 4z â€“ 12 = z2Â â€“ 4z + 22Â â€“ 22Â â€“ 12 [Adding and subtracting 22]

= (z â€“ 2)2Â â€“ 42Â (Completing the square)

By using the formulaÂ (a2Â â€“ b2) = (a+b) (a-b)

= (z â€“ 2 + 4) (z â€“ 2 â€“ 4)

= (z â€“ 6) (z + 2)