**RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization** is the best study material for students who are finding difficulties in solving problems. RD Sharma Solutions for Class 8 Maths includes answers to all the questions provided in the textbook that is prescribed for Class 8 in accordance with the CBSE Board. The subject experts at BYJU’S have formulated the solutions with utmost care to help students secure good marks in their annual examinations. The PDF of this chapter is available here, which students can download for free from the links given below. Chapter 7 – Factorization contains nine exercises, and the RD Sharma Solutions available on 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.

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

### Access Answers to 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. 2x ^{2} and 12x^{2}**

**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 the given monomial is x

The smallest power of x in two monomials is 2

The monomial of common literals with the smallest power is x^{2}

∴ The greatest common factor = 2x^{2}

**2. 6x ^{3}y and 18x^{2}y^{3}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in three monomials is 2

The smallest power of y in three monomials is 1

The Monomial of common literals with the smallest power is x^{2}y

∴ The greatest common factor = 6x^{2}y

**3. 7x, 21x ^{2} and 14xy^{2}**

**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

The smallest power of x in three monomials is 1

The smallest power of y in three monomials is 0

Monomial of common literals with the smallest power is x

∴ The greatest common factor = 7x

**4. 42x ^{2}yz and 63x^{3}y^{2}z^{3}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in two monomials is 2

The smallest power of y in two monomials is 1

The smallest power of z in two monomials is 1

The monomial of common literals with the smallest power is x^{2}yz

∴ The greatest common factor = 21x^{2}yz

**5. 12ax ^{2}, 6a^{2}x^{3} and 2a^{3}x^{5}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in three monomials is 2

The smallest power of a in three monomials is 1

The monomial of common literals with the smallest power is ax^{2}

∴ The greatest common factor = 2ax^{2}

**6. 9x ^{2}, 15x^{2}y^{3}, 6xy^{2} and 21x^{2}y^{2}**

**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

The smallest power of x in four monomials is 1

The smallest power of y in four monomials is 0

The monomials of common literals with the smallest power is x

∴ The greatest common factor = 3x

**7. 4a ^{2}b^{3}, -12a^{3}b, 18a^{4}b^{3}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of a in three monomials is 2

The smallest power of b in three monomials is 1

The monomials of common literals with the smallest power is a^{2}b

∴ The greatest common factor = 2a^{2}b

**8. 6x ^{2}y^{2}, 9xy^{3}, 3x^{3}y^{2}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in three monomials is 1

The smallest power of y in three monomials is 2

The monomials of common literals with the smallest power is xy^{2}

∴ The greatest common factor = 3xy^{2}

**9. a ^{2}b^{3}, a^{3}b^{2}**

**Solution:**

We know that the numerical coefficients of the given numerical are 0

Common literals appearing in given numerical are a and b

The smallest power of a in two monomials = 2

The smallest power of b in two monomials = 2

The monomials of common literals with the smallest power is a^{2}b^{2}

∴ The greatest common factor = a^{2}b^{2}

**10. 36a ^{2}b^{2}c^{4}, 54a^{5}c^{2}, 90a^{4}b^{2}c^{2}**

**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 the given numerical are a, b and c

The smallest power of a in three monomials is 2

The smallest power of b in three monomials is 0

The smallest power of c in three monomials is 2

The monomials of common literals with the smallest power is a^{2}c^{2}

∴ The greatest common factor = 18a^{2}c^{2}

**11. x ^{3}, -yx^{2}**

**Solution:**

We know that the numerical coefficients of the given numerical are 0

Common literals appearing in given numerical are x and y

The smallest power of x in two monomials is 2

The smallest power of y in two monomials is 0

The monomials of common literals with the smallest power is x^{2}

∴ The greatest common factor = x^{2}

**12. 15a ^{3}, -45a^{2}, -150a**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of a in three monomials is 1

The monomials of common literals with the smallest power is a

∴ The greatest common factor = 15a

**13. 2x ^{3}y^{2}, 10x^{2}y^{3}, 14xy**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in three monomials is 1

The smallest power of y in three monomials is 1

The monomials of common literals with the smallest power is xy

∴ The greatest common factor = 2xy

**14. 14x ^{3}y^{5}, 10x^{5}y^{3}, 2x^{2}y^{2}**

**Solution:**

We know that the numerical coefficients of the 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

The smallest power of x in three monomials is 2

The smallest power of y in three monomials is 2

The monomials of common literals with the smallest power is x^{2}y^{2}

∴ The greatest common factor = 2x^{2}y^{2}

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

**15. 5a ^{4} + 10a^{3} – 15a^{2}**

**Solution:**

The greatest common factor of the three terms is 5a^{2}

**16. 2xyz + 3x ^{2}y + 4y^{2}**

**Solution:**

The greatest common factor of the three terms is y

**17. 3a ^{2}b^{2} + 4b^{2}c^{2} + 12a^{2}b^{2}c^{2}**

**Solution**:

The greatest common factor of the three terms is b^{2}.

### EXERCISE 7.2 PAGE NO: 7.5

**Factorize the following:**

**1. 3x – 9**

**Solution:**

The greatest common factor in the given two terms is 3

3x – 9

3 (x – 3)

**2. 5x – 15x ^{2}**

**Solution:**

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

5x – 15x^{2}

5x (1 – 3x)

**3. 20a ^{12}b^{2} – 15a^{8}b^{4}**

**Solution:**

Greatest common factor in the given two terms is 5a^{8}b^{2}

20a^{12}b^{2} – 15a^{8}b^{4}

5a^{8}b^{2} (4a^{4} – 3b^{2})

**4. 72x ^{6}y^{7} – 96x^{7}y^{6}**

Solution:

Greatest common factor in the given two terms is 24x^{6}y^{6}

72x^{6}y^{7} – 96x^{7}y^{6}

24x^{6}y^{6} (3y – 4x)

**5. 20x ^{3} – 40x^{2} + 80x**

**Solution:**

Greatest common factor in the given three terms is 20x

20x^{3} – 40x^{2} + 80x

20x (x^{2} – 2x +4)

**6. 2x ^{3}y^{2} – 4x^{2}y^{3} + 8xy^{4}**

**Solution:**

Greatest common factor in the given three terms is 2xy^{2}

2x^{3}y^{2} – 4x^{2}y^{3} + 8xy^{4}

2xy^{2} (x^{2} – 2xy + 4y^{2})

**7. 10m ^{3}n^{2} + 15m^{4}n – 20m^{2}n^{3}**

**Solution:**

Greatest common factor in the given three terms is 5mn^{2}

10m^{3}n^{2} + 15m^{4}n – 20m^{2}n^{3}

5m^{2}n (2mn + 3m^{2} – 4n^{2})

**8. 2a ^{4}b^{4} – 3a^{3}b^{5} + 4a^{2}b^{5}**

**Solution:**

Greatest common factor in the given three terms is a^{2}b^{4}

2a^{4}b^{4} – 3a^{3}b^{5} + 4a^{2}b^{5}

a^{2}b^{4} (2a^{2} – 3ab + 4b)

**9. 28a ^{2} + 14a^{2}b^{2} – 21a^{4}**

**Solution:**

Greatest common factor in the given three terms is 7a^{2}

28a^{2} + 14a^{2}b^{2} – 21a^{4}

7a^{2 }(4a + 2b^{2} – 3a^{2})

**10. a ^{4}b – 3a^{2}b^{2} – 6ab^{3}**

**Solution:**

Greatest common factor in the given three terms is ab

a^{4}b – 3a^{2}b^{2} – 6ab^{3}

ab (a^{3} – 3ab – 6b^{2})

**11. 2l ^{2}mn – 3lm^{2}n + 4lmn^{2}**

**Solution:**

Greatest common factor in the given three terms is lmn

2l^{2}mn – 3lm^{2}n + 4lmn^{2}

lmn (2l – 3m + 4n)

**12. x ^{4}y^{2} – x^{2}y^{4} – x^{4}y^{4}**

**Solution:**

Greatest common factor in the given three terms is x^{2}y^{2}

x^{4}y^{2} – x^{2}y^{4} – x^{4}y^{4}

x^{2}y^{2} (x^{2} – y^{2} – x^{2}y^{2})

**13. 9x ^{2}y + 3axy**

**Solution:**

Greatest common factor in the given three terms is 3xy

9x^{2}y + 3axy

3xy (3x + a)

**14. 16m – 4m ^{2}**

**Solution:**

Greatest common factor in the given two terms is 4m

16m – 4m^{2}

4m (4 – m)

**15. -4a ^{2} + 4ab – 4ca**

**Solution:**

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

-4a^{2} + 4ab – 4ca

-4a (a – b + c)

**16. **x^{2}yz** + xy ^{2}z + xyz^{2}**

**Solution:**

Greatest common factor in the given three terms is xyz

x^{2}yz + xy^{2}z + xyz^{2}

xyz (x + y +z)

**17. ax ^{2}y + bxy^{2} + cxyz**

**Solution:**

Greatest common factor in the given three terms is xy

ax^{2}y + bxy^{2} + 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 – y) as common, we get,

(6x + 7y) (2x – 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 – y) + s (x – y)

By taking (x – y) as common, we get,

(x – y) (-2r + s)

(x – y) (s – 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 – 3)

**4. 9a (6a – 5b) – 12a ^{2} (6a – 5b)**

**Solution:**

We have,

9a (6a – 5b) – 12a^{2} (6a – 5b)

By taking (6a – 5b) as common, we get,

(9a – 12a^{2}) (6a – 5b)

3a(3 – 4a) (6a – 5b)

**5. 5 (x – 2y) ^{2} + 3 (x – 2y)**

**Solution:**

We have,

5 (x – 2y)^{2} + 3 (x – 2y)

By taking (x – 2y) as common, we get,

(x – 2y) [5 (x – 2y) + 3]

(x – 2y) (5x – 10y + 3)

**6. 16 (2l – 3m) ^{2} – 12 (3m – 2l)**

**Solution:**

We have,

16 (2l – 3m)^{2} – 12 (3m – 2l)

By taking (-1) as common, we get,

16 (2l – 3m)^{2} + 12 (2l – 3m)

By taking 4(2l – 3m) as common we get,

4(2l – 3m) [4 (2l – 3m) + 3]

4(2l – 3m) (8l – 12m + 3)

**7. 3a (x – 2y) – b (x – 2y)**

**Solution:**

We have,

3a (x – 2y) – b (x – 2y)

By taking (x – 2y) as common, we get,

(3a – b) (x – 2y)

**8. a ^{2} (x + y) + b^{2} (x + y) + c^{2} (x + y)**

**Solution:**

We have,

a^{2} (x + y) + b^{2} (x + y) + c^{2} (x + y)

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

(a^{2} + b^{2} + c^{2}) (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 – y) (x – y + 1)

**10. 6 (a + 2b) – 4 (a + 2b) ^{2}**

**Solution:**

We have,

6 (a + 2b) – 4 (a + 2b)^{2}

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

[6 – 4 (a + 2b)] (a + 2b)(6 – 4a – 8b) (a + 2b)

2(3 – 2a – 4b) (a + 2b)

**11. a (x – y) + 2b (y – x) + c (x – y) ^{2}**

**Solution:**

We have,

a (x – y) + 2b (y – x) + c (x – y)^{2}

By taking (-1) as common, we get,

a (x – y) – 2b (x – y) + c (x – y)^{2}

By taking (x – y) as common, we get,

[a – 2b + c(x – y)] (x – y)(x – y) (a – 2b + cx – cy)

**12. -4 (x – 2y) ^{2} + 8 (x – 2y)**

**Solution:**

We have,

-4 (x – 2y)^{2} + 8 (x – 2y)

By taking 4(x – 2y) as common, we get,

[-(x – 2y) + 2] 4(x – 2y)4(x – 2y) (-x + 2y + 2)

**13. x ^{3} (a – 2b) + x^{2} (a – 2b)**

**Solution:**

We have,

x^{3} (a – 2b) + x^{2} (a – 2b)

By taking x^{2} (a – 2b) as common, we get,

(x + 1) [x^{2} (a – 2b)]

x^{2} (a – 2b) (x + 1)

**14. (2x – 3y) (a + b) + (3x – 2y) (a + b)**

**Solution:**

We have,

(2x – 3y) (a + b) + (3x – 2y) (a + b)

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

(a + b) [(2x – 3y) + (3x – 2y)]

(a + b) [2x -3y + 3x – 2y]

(a + b) [5x – 5y]

(a + b) 5(x – y)

**15. 4(x + y) (3a – b) + 6(x + y) (2b – 3a)**

**Solution:**

We have,

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

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

(x + y) [4(3a – b) + 6(2b – 3a)]

(x + y) [12a – 4b + 12b – 18a]

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

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

(x + y) 2(4b – 3a)

### EXERCISE 7.4 PAGE NO: 7.12

**Factorize each of the following expressions:**

**1. qr – pr + qs – ps**

**Solution:**

We have,

qr – pr + qs – ps

By grouping similar terms, we get,

qr + qs – pr – ps

q(r + s) –p (r + s)

(q – p) (r + s)

**2. p ^{2}q – pr^{2} – pq + r^{2}**

**Solution:**

We have,

p^{2}q – pr^{2} – pq + r^{2}

By grouping similar terms, we get,

p^{2}q – pq – pr^{2} + r^{2}

pq(p – 1) –r^{2}(p – 1)

(p – 1) (pq – r^{2})

**3. 1 + x + xy + x ^{2}y**

**Solution:**

We have,

1 + x + xy + x^{2}y

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

(1 + x) (1 + xy)

**4. ax + ay – bx – by**

**Solution:**

We have,

ax + ay – bx – by

a(x + y) –b (x + y)

(a – b) (x + y)

**5. xa ^{2} + xb^{2} – ya^{2} – yb^{2}**

**Solution:**

We have,

xa^{2} + xb^{2} – ya^{2} – yb^{2}

x(a^{2} + b^{2}) –y (a^{2} + b^{2})

(x – y) (a^{2} + b^{2})

**6. x ^{2} + xy + xz + yz**

**Solution:**

We have,

x^{2} + 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 – by – ay + y ^{2}**

**Solution:**

We have,

ab – by – ay + y^{2}

By grouping similar terms, we get,

Ab – ay – by + y^{2}

a (b – y) – y (b – y)

(a – y) (b – y)

**9. axy + bcxy – az – bcz**

**Solution:**

We have,

axy + bcxy – az – bcz

By grouping similar terms, we get,

axy – az + bcxy – bcz

a (xy – z) + bc (xy – z)

(a + bc) (xy – z)

**10. lm ^{2} – mn^{2} – lm + n^{2}**

**Solution:**

We have,

lm^{2} – mn^{2} – lm + n^{2}

By grouping similar terms, we get,

lm^{2} – lm – mn^{2} + n^{2}

lm (m – 1) – n^{2} (m – 1)

(lm – n^{2}) (m – 1)

**11. x ^{3} – y^{2} + x – x^{2}y^{2}**

**Solution:**

We have,

x^{3} – y^{2} + x – x^{2}y^{2}

By grouping similar terms, we get,

x + x^{3} – y^{2} – x^{2}y^{2}

x (1 + x^{2}) – y^{2} (1 + x^{2})

(x – y^{2}) (1 + x^{2})

**12. 6xy + 6 – 9y – 4x**

**Solution:**

We have,

6xy + 6 – 9y – 4x

By grouping similar terms, we get,

6xy – 4x – 9y + 6

2x (3y – 2) – 3 (3y – 2)

(2x – 3) (3y – 2)

**13. x ^{2} – 2ax – 2ab + bx**

**Solution:**

We have,

x^{2} – 2ax – 2ab + bx

By grouping similar terms, we get,

x^{2} + bx – 2ax – 2ab

x (x + b) – 2a (x + b)

(x – 2a) (x + b)

**14. x ^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}**

**Solution:**

We have,

x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

By grouping similar terms, we get,

x^{3} + 3xy^{2} – 2x^{2}y – 6y^{3}

x (x^{2} + 3y^{2}) – 2y (x^{2} + 3y^{2})

(x – 2y) (x^{2} + 3y^{2})

**15. abx ^{2} + (ay – b) x – y**

**Solution:**

We have,

abx^{2} + (ay – b) x – y

abx^{2} + ayx – bx – y

By grouping similar terms, we get,

abx^{2} – bx + ayx – y

bx (ax – 1) + y (ax – 1)

(bx + y) (ax – 1)

**16. (ax + by) ^{2} + (bx – ay)^{2}**

**Solution:**

We have,

(ax + by)^{2} + (bx – ay)^{2}

a^{2}x^{2} + b^{2}y^{2} + 2axby + b^{2}x^{2} + a^{2}y^{2} – 2axby

a^{2}x^{2} + b^{2}y^{2} + b^{2}x^{2} + a^{2}y^{2}

By grouping similar terms, we get,

a^{2}x^{2} + a^{2}y^{2} + b^{2}y^{2} + b^{2}x^{2}

a^{2} (x^{2} + y^{2}) + b^{2} (x^{2} + y^{2})

(a^{2} + b^{2}) (x^{2} + y^{2})

**17. 16 (a – b) ^{3} – 24 (a – b)^{2}**

**Solution:**

We have,

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 (x ^{2} + 1) + x (a^{2} + b^{2})**

**Solution:**

We have,

ab(x^{2} + 1) + x(a^{2} + b^{2})

abx^{2} + ab + xa^{2} + xb^{2}

By grouping similar terms, we get,

abx^{2} + xa^{2} + xb^{2} + ab

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

(ax + b) (bx + a)

**19. a ^{2}x^{2} + (ax^{2} + 1)x + a**

**Solution:**

We have,

a^{2}x^{2} + (ax^{2} + 1)x + a

a^{2}x^{2} + ax^{3} + x + a

ax^{2} (a + x) + 1 (x + a)

(x + a) (ax^{2} + 1)

**20. a (a – 2b – c) + 2bc**

**Solution:**

We have,

a (a – 2b – c) + 2bc

a^{2} – 2ab – ac + 2bc

a (a – 2b) – c (a – 2b)

(a – 2b) (a – c)

**21. a (a + b – c) – bc**

**Solution:**

We have,

a (a + b – c) – bc

a^{2} + ab – ac – bc

a (a + b) – c (a + b)

(a + b) (a – c)

**22. x ^{2} – 11xy – x + 11y**

**Solution:**

We have,

x^{2} – 11xy – x + 11y

By grouping similar terms, we get,

x^{2} – x – 11xy + 11y

x (x – 1) – 11y (x – 1)

(x – 11y) (x – 1)

**23. ab – a – b + 1**

**Solution:**

We have,

ab – a – b + 1

a (b – 1) – 1 (b – 1)

(a – 1) (b – 1)

**24. x ^{2} + y – xy – x**

**Solution:**

We have,

x^{2} + y – xy – x

By grouping similar terms, we get,

x^{2} – x + y – xy

x (x – 1) – y (x – 1)

(x – y) (x – 1)

### EXERCISE 7.5 PAGE NO: 7.17

**Factorize each of the following expressions:**

**1. 16x ^{2} – 25y^{2}**

**Solution:**

We have,

16x^{2} – 25y^{2}

(4x)^{2} – (5y)^{2}

By using the formula (a^{2} – b^{2}) = (a + b) (a – b) we get,

(4x + 5y) (4x – 5y)

**2. 27x ^{2} – 12y^{2}**

**Solution:**

We have,

27x^{2} – 12y^{2}

By taking 3 as common, we get,

3 [(3x)^{2} – (2y)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

3 (3x + 2y) (3x – 2y)

**3. 144a ^{2} – 289b^{2}**

**Solution:**

We have,

144a^{2} – 289b^{2}

(12a)^{2} – (17b)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(12a + 17b) (12a – 17b)

**4. 12m ^{2} – 27**

**Solution:**

We have,

12m^{2} – 27

By taking 3 as common, we get,

3 (4m^{2} – 9)

3 [(2m)^{2} – 3^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

3 (2m + 3) (2m – 3)

**5. 125x ^{2} – 45y^{2}**

**Solution:**

We have,

125x^{2} – 45y^{2}

By taking 5 as common, we get,

5 (25x^{2} – 9y^{2})

5 [(5x)^{2} – (3y)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

5 (5x + 3y) (5x – 3y)

**6. 144a ^{2} – 169b^{2}**

**Solution:**

We have,

144a^{2} – 169b^{2}

(12a)^{2} – (13b)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(12a + 13b) (12a – 13b)

**7. (2a – b) ^{2} – 16c^{2}**

**Solution:**

We have,

(2a – b)^{2} – 16c^{2}

(2a – b)^{2} – (4c)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(2a – b + 4c) (2a – b – 4c)

**8. (x + 2y) ^{2} – 4 (2x – y)^{2}**

**Solution:**

We have,

(x + 2y)^{2} – 4 (2x – y)^{2}

(x + 2y)^{2} – [2 (2x – y)]^{2}

By using the formula (a^{2} – b^{2})= (a + b) (a – b) we get,

(x + 4x + 2y – 2y) (x – 4x + 2y + 2y)

(5x) (4y – 3x)

**9. 3a ^{5} – 48a^{3}**

**Solution:**

We have,

3a^{5} – 48a^{3}

By taking 3 as common, we get,

3a^{3} (a^{2} – 16)

3a^{3} (a^{2} – 4^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

3a^{3} (a + 4) (a – 4)

**10. a ^{4} – 16b^{4}**

**Solution:**

We have,

a^{4} – 16b^{4}

(a^{2})^{2} – (4b^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2} + 4b^{2}) (a^{2} – 4b^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2} + 4b^{2}) (a + 2b) (a – 2b)

**11. x ^{8} – 1**

**Solution:**

We have,

x^{8} – 1

(x^{4})^{2}–(1)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{4} + 1) (x^{4} – 1)

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{4} + 1) (x^{2} + 1) (x – 1) (x + 1)

**12. 64 – (a + 1) ^{2}**

**Solution:**

We have,

64 – (a + 1)^{2}

8^{2} – (a + 1)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a + 9) (7 – a)

**13. 36l ^{2} – (m + n)^{2}**

**Solution:**

We have,

36l^{2} – (m + n)^{2}

(6l)^{2} – (m + n)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(6l + m + n) (6l – m – n)

**14. 25x ^{4}y^{4} – 1**

**Solution:**

We have,

25x^{4}y^{4} – 1

(5x^{2}y^{2})^{2} – (1)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(5x^{2}y^{2} – 1) (5x^{2}y^{2} + 1)

**15. a ^{4} – 1/b^{4}**

**Solution:**

We have,

a^{4} – 1/b^{4}

(a^{2})^{2} – (1/b^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2} + 1/b^{2}) (a^{2} – 1/b^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2} + 1/b^{2}) (a – 1/b) (a + 1/b)

**16. x ^{3} – 144x**

**Solution:**

We have,

x^{3} – 144x

x [x^{2} – (12)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

x (x + 12) (x – 12)

**17. (x – 4y) ^{2} – 625**

**Solution:**

We have,

(x – 4y)^{2} – 625

(x – 4y)^{2} – (25)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x – 4y + 25) (x – 4y – 25)

**18. 9 (a – b) ^{2} – 100 (x – y)^{2}**

**Solution:**

We have,

9 (a – b)^{2} – 100 (x – y)^{2}

^{2}– [10 (x – y)]

^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

**19. (3 + 2a) ^{2} – 25a^{2}**

**Solution:**

We have,

(3 + 2a)^{2} – 25a^{2}

(3 + 2a)^{2} – (5a)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(3 + 2a + 5a) (3 + 2a – 5a)

(3 + 7a) (3 – 3a)

(3 + 7a) 3(1 – a)

**20. (x + y) ^{2} – (a – b)^{2}**

**Solution:**

We have,

(x + y)^{2} – (a – b)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x + y + a – b) (x + y – a + b)

**21. 1/16x ^{2}y^{2} – 4/49y^{2}z^{2}**

**Solution:**

We have,

1/16x^{2}y^{2} – 4/49y^{2}z^{2}

(1/4xy)^{2} – (2/7yz)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(xy/4 + 2yz/7) (xy/4 – 2yz/7)

y^{2} (x/4 + 2/7z) (x/4 – 2/7z)

**22. 75a ^{3}b^{2} – 108ab^{4}**

**Solution:**

We have,

75a^{3}b^{2} – 108ab^{4}

3ab^{2} (25a^{2} – 36b^{2})

3ab^{2} [(5a)^{2} – (6b)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

3ab^{2} (5a + 6b) (5a – 6b)

**23. x ^{5} – 16x^{3}**

**Solution:**

We have,

x^{5} – 16x^{3}

x^{3} (x^{2} – 16)

x^{3} (x^{2} – 4^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

x^{3} (x + 4) (x – 4)

**24. 50/x ^{2} – 2x^{2}/81**

**Solution:**

We have,

50/x^{2} – 2x^{2}/81

2 (25/x^{2} – x^{2}/81)

2 [(5/x)^{2} – (x/9)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

2 (5/x+ x/9) (5/x – x/9)

**25. 256x ^{3} – 81x**

**Solution:**

We have,

256x^{3} – 81x

x (256x^{4} – 81)

x [(16x^{2})^{2} – 9^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

x (4x + 3) (4x – 3) (16x^{2} + 9)

**26. a ^{4} – (2b + c)^{4}**

**Solution:**

We have,

a^{4} – (2b + c)^{4}

(a^{2})^{2} – [(2b + c)^{2}]^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

^{2}+ (2b + c)

^{2}] [a

^{2}– (2b + c)

^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

^{2}+ (2b + c)

^{2}] [a + 2b + c] [a – 2b – c]

**27. (3x + 4y) ^{4} – x^{4}**

**Solution:**

We have,

(3x + 4y)^{4} – x^{4}

^{2}]

^{2}– (x

^{2})

^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

^{2}+ x

^{2}] [(3x + 4y)

^{2}– x

^{2}] [(3x + 4y)

^{2}+ x

^{2}] [3x + 4y + x] [3x + 4y – x] [(3x + 4y)

^{2}+ x

^{2}] [4x + 4y] [2x + 4y] [(3x + 4y)

^{2}+ x

^{2}] 8[x + 2y] [x + y]

**28. p ^{2}q^{2} – p^{4}q^{4}**

**Solution:**

We have,

p^{2}q^{2} – p^{4}q^{4}

(pq)^{2} – (p^{2}q^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(pq + p^{2}q^{2}) (pq – p^{2}q^{2})

p^{2}q^{2} (1 + pq) (1 – pq)

**29. 3x ^{3}y – 243xy^{3}**

**Solution:**

We have,

3x^{3}y – 243xy^{3}

3xy (x^{2} – 81y^{2})

3xy [x^{2} – (9y)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(3xy) (x + 9y) (x – 9y)

**30. a ^{4}b^{4} – 16c^{4}**

**Solution:**

We have,

a^{4}b^{4} – 16c^{4}

(a^{2}b^{2})^{2} – (4c^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2}b^{2} + 4c^{2}) (a^{2}b^{2} – 4c^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2}b^{2} + 4c^{2}) (ab + 2c) (ab – 2c)

**31. x ^{4} – 625**

**Solution:**

We have,

x^{4} – 625

(x^{2})^{2} – (25)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{2} + 25) (x^{2} – 25)

(x^{2} + 25) (x^{2} – 5^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{2} + 25) (x + 5) (x – 5)

**32. x ^{4} – 1**

**Solution:**

We have,

x^{4} – 1

(x^{2})^{2} – (1)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{2} + 1) (x^{2} – 1)

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{2} + 1) (x + 1) (x – 1)

**33. 49(a – b) ^{2} – 25(a + b)^{2}**

**Solution:**

We have,

49(a – b)^{2} – 25(a + b)^{2}

^{2}– [5 (a + b)]

^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(7a – 7b + 5a + 5b) (7a – 7b – 5a – 5b)

(12a – 2b) (2a – 12b)

2 (6a – b) 2 (a – 6b)

4 (6a – b) (a – 6b)

**34. x – y – x ^{2} + y^{2}**

**Solution:**

We have,

x – y – x^{2} + y^{2}

x – y – (x^{2} – y^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

x – y – (x + y) (x – y)

(x – y) (1 – x – y)

**35. 16(2x – 1) ^{2} – 25y^{2}**

**Solution:**

We have,

16(2x – 1)^{2} – 25y^{2}

^{2}– (5y)

^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(8x + 5y – 4) (8x – 5y – 4)

**36. 4(xy + 1) ^{2} – 9(x – 1)^{2}**

**Solution:**

We have,

4(xy + 1)^{2} – 9(x – 1)^{2}

^{2}– [3 (x – 1)]

^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)

(2xy + 3x – 1) (2xy – 3x + 5)

**37. (2x + 1) ^{2} – 9x^{4}**

**Solution:**

We have,

(2x + 1)^{2} – 9x^{4}

(2x + 1)^{2} – (3x^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(2x + 1 + 3x^{2}) (2x + 1 – 3x^{2})

(3x^{2} + 2x + 1) (-3x^{2} + 2x + 1)

**38. x ^{4} – (2y – 3z)^{2}**

**Solution:**

We have,

x^{4} – (2y – 3z)^{2}

(x^{2})^{2} – (2y – 3z)^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(x^{2} + 2y – 3z) (x^{2} – 2y + 3z)

**39. a ^{2 }– b^{2} + a – b**

**Solution:**

We have,

a^{2} – b^{2} + a – b

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a + b) (a – b) + (a – b)

(a – b) (a + b + 1)

**40. 16a ^{4} – b^{4}**

**Solution:**

We have,

16a^{4} – b^{4}

(4a^{2})^{2} – (b^{2})^{2}

(4a^{2} + b^{2}) (4a^{2} – b^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(4a^{2} + b^{2}) (2a + b) (2a – b)

**41. a ^{4} – 16(b – c)^{4}**

**Solution:**

We have,

a^{4} – 16(b – c)^{4}

(a^{2})^{2} – [4 (b – c)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

^{2}+ 4 (b – c)

^{2}] [a

^{2}– 4 (b – c)

^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

^{2}+ 4 (b – c)

^{2}] [(a + 2b – 2c) (a – 2b + 2c)]

**42. 2a ^{5} – 32a**

**Solution:**

We have,

2a^{5} – 32a

2a (a^{4} – 16)

2a [(a^{2})^{2} – (4)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

2a (a^{2} + 4) (a^{2} – 4)

2a (a^{2} + 4) (a^{2} – 2^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

2a (a^{2} + 4) (a + 2) (a – 2)

**43. a ^{4}b^{4} – 81c^{4}**

**Solution:**

We have,

a^{4}b^{4} – 81c^{4}

(a^{2}b^{2})^{2} – (9c^{2})^{2}

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2}b^{2} + 9c^{2}) (a^{2}b^{2} – 9c^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

(a^{2}b^{2} + 9c^{2}) (ab + 3c) (ab – 3c)

**44. xy ^{9} – yx^{9}**

**Solution:**

We have,

xy^{9} – yx^{9}

-xy (x^{8} – y^{8})

-xy [(x^{4})^{2} – (y^{4})^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

-xy (x^{4} + y^{4}) (x^{4} – y^{4})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

-xy (x^{4} + y^{4}) (x^{2} + y^{2}) (x^{2} – y^{2})

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

-xy (x^{4} + y^{4}) (x^{2} + y^{2}) (x + y) (x – y)

**45. x ^{3} – x**

**Solution:**

We have,

x^{3} – x

x (x^{2} – 1)

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

x (x + 1) (x – 1)

**46. 18a ^{2}x^{2} – 32**

**Solution:**

We have,

18a^{2}x^{2} – 32

2 [(3ax)^{2} – (4)^{2}]

By using the formula (a^{2} – b^{2}) = (a-b) (a+b)

2 (3ax + 4) (3ax – 4)

### EXERCISE 7.6 PAGE NO: 7.22

**Factorize each of the following algebraic expressions:**

**1. 4x ^{2} + 12xy + 9y^{2}**

**Solution:**

We have,

4x^{2} + 12xy + 9y^{2}

By using the formula (x + y)^{2} = x^{2} + y^{2} + 2xy

(2x)^{2} + (3y)^{2} + 2 (2x) (3y)

(2x + 3y)^{2}

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

**2. 9a ^{2} – 24ab + 16b^{2}**

**Solution:**

We have,

9a^{2} – 24ab + 16b^{2}

By using the formula (x – y)^{2} = x^{2} + y^{2} – 2xy

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

(3a)^{2} + (4b)^{2} – 2 (3a) (4b)

(3a – 4b)^{2}

(3a – 4b) (3a – 4b)

**3. p ^{2}q^{2} – 6pqr + 9r^{2}**

**Solution:**

We have,

p^{2}q^{2} – 6pqr + 9r^{2}

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(pq)^{2} + (3r)^{2} – 2 (pq) (3r)

(pq – 3r)^{2}

(pq – 3r) (pq – 3r)

**4. 36a ^{2} + 36a + 9**

**Solution:**

We have,

36a^{2} + 36a + 9

(6a)^{2} + 2 × 6a × 3 + 3^{2}

(6a + 3)^{2}

**5. a ^{2} + 2ab + b^{2} – 16**

**Solution:**

We have,

a^{2} + 2ab + b^{2} – 16

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(a + b)^{2} – 4^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a + b + 4) (a + b – 4)

**6. 9z ^{2} – x^{2} + 4xy – 4y^{2}**

**Solution:**

We have,

9z^{2} – x^{2} + 4xy – 4y^{2}

(3z)^{2} – [x^{2} – 2 (x) (2y) + (2y)^{2}]

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(3z)^{2} – (x – 2y)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

**7. 9a ^{4} – 24a^{2}b^{2} + 16b^{4} – 256**

**Solution:**

We have,

9a^{4} – 24a^{2}b^{2} + 16b^{4} – 256

(3a^{2})^{2} – 2 (4a^{2}) (3b^{2}) + (4b^{2})^{2} – (16)^{2}

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(3a^{2} – 4b^{2})^{2} – (16)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(3a^{2} – 4b^{2} + 16) (3a^{2} – 4b^{2} – 16)

**8. 16 – a ^{6} + 4a^{3}b^{3} – 4b^{6}**

**Solution:**

We have,

16 – a^{6} + 4a^{3}b^{3} – 4b^{6}

4^{2} – [(a^{3})^{2} – 2 (a^{3}) (2b^{3}) + (2b^{3})^{2}]

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

4^{2} – (a^{3} – 2b^{3})^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

^{3}– 2b

^{3})] [4 – (a

^{3}– 2b

^{3})]

**9. a ^{2} – 2ab + b^{2} – c^{2}**

**Solution:**

We have,

a^{2} – 2ab + b^{2} – c^{2}

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(a – b)^{2} – c^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a – b + c) (a – b – c)

**10. x ^{2} + 2x + 1 – 9y^{2}**

**Solution:**

We have,

x^{2} + 2x + 1 – 9y^{2}

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(x + 1)^{2} – (3y)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(x + 3y + 1) (x – 3y + 1)

**11. a ^{2} + 4ab + 3b^{2}**

**Solution:**

We have,

a^{2} + 4ab + 3b^{2}

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

a^{2} + ab + 3ab + 3b^{2}

By grouping we get,

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

(a + 3b) (a + b)

**12. 96 – 4x – x ^{2}**

**Solution:**

We have,

96 – 4x – x^{2}

-x^{2} – 4x + 96

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

-x^{2} – 12x + 8x + 96

By grouping, we get,

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

(x + 12) (-x + 8)

**13. a ^{4} + 3a^{2} + 4**

**Solution:**

We have,

a^{4} + 3a^{2} + 4

(a^{2})^{2} + (a^{2})^{2} + 2 (2a^{2}) + 4 – a^{2}

(a^{2} + 2)^{2} + (-a^{2})

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a^{2} + 2 + a) (a^{2} + 2 – a)

(a^{2} + a + 2) (a^{2} – a + 2)

**14. 4x ^{4} + 1**

**Solution:**

We have,

4x^{4} + 1

(2x^{2})^{2} + 1 + 4x^{2} – 4x^{2}

(2x^{2} + 1)^{2} – 4x^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(2x^{2} + 1 + 2x) (2x^{2} + 1 – 2x)

(2x^{2} + 2x + 1) (2x^{2} – 2x + 1)

**15. 4x ^{4} + y^{4}**

**Solution:**

We have,

4x^{4} + y^{4}

(2x^{2})^{2} + (y^{2})^{2} + 4x^{2}y^{2} – 4x^{2}y^{2}

(2x^{2} + y^{2})^{2} – 4x^{2}y^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(2x^{2} + y^{2} + 2xy) (2x^{2} + y^{2} – 2xy)

**16. (x + 2) ^{4} – 6(x + 2) + 9**

**Solution**:

We have,

(x + 2)^{4} – 6(x + 2) + 9

(x^{2} + 2^{2})^{2} – 6x – 12 + 9

(x^{2} + 2^{2} + 2(2)(x)) – 6x – 12 + 9

x^{2} + 4 + 4x – 6x – 12 + 9

x^{2} – 2x + 1

By using the formula (a – b)^{2} = a^{2} + b^{2} – 2ab

(x – 1)^{2}

**17. 25 – p ^{2} – q^{2} – 2pq**

**Solution:**

We have,

25 – p^{2} – q^{2} – 2pq

25 – (p^{2} + q^{2} + 2pq)

(5)^{2} – (p + q)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(5 + p + q) (5 –p – q)

-(p + q + 5) (p + q – 5)

**18. x ^{2} + 9y^{2} – 6xy – 25a^{2}**

**Solution:**

We have,

x^{2} + 9y^{2} – 6xy – 25a^{2}

(x – 3y)^{2} – (5a)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(x – 3y + 5a) (x – 3y – 5a)

**19. 49 – a ^{2} + 8ab – 16b^{2}**

**Solution:**

We have,

49 – a^{2} + 8ab – 16b^{2}

49 – (a^{2} – 8ab + 16b^{2})

49 – (a – 4b)^{2}

By using the formula (a^{2} – b^{2}) = (a + b) (a – b)

(7 + a – 4b) (7 – a + 4b)

-(a – 4b + 7) (a – 4b – 7)

**20. a ^{2} – 8ab + 16b^{2} – 25c^{2} **

**Solution:**

We have,

a^{2} – 8ab + 16b^{2} – 25c^{2}

(a – 4b)^{2}– (5c)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a – 4b + 5c) (a – 4b – 5c)

**21. x ^{2} – y^{2} + 6y – 9 **

**Solution:**

We have,

x^{2} – y^{2} + 6y – 9

x^{2} + 6y – (y^{2} – 6y + 9)

x^{2} – (y – 3)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(x + y – 3) (x – y + 3)

**22. 25x ^{2} – 10x + 1 – 36y^{2} **

**Solution:**

We have,

25x^{2} – 10x + 1 – 36y^{2}

(5x)^{2} – 2 (5x) + 1 – (6y)^{2}

(5x – 1)^{2} – (6y)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(5x – 6y – 1) (5x + 6y – 1)

**23. a ^{2} – b^{2} + 2bc – c^{2} **

**Solution:**

We have,

a^{2} – b^{2} + 2bc – c^{2}

a^{2} – (b^{2} – 2bc + c^{2})

a^{2} – (b – c)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a + b – c) (a – b + c)

**24. a ^{2} + 2ab + b^{2} – c^{2}**

**Solution:**

We have,

a^{2} + 2ab + b^{2} – c^{2}

(a + b)^{2} – c^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a + b + c) (a + b – c)

**25. 49 – x ^{2} – y^{2} + 2xy**

**Solution:**

We have,

49 – x^{2} – y^{2} + 2xy

49 – (x^{2} + y^{2} – 2xy)

7^{2} – (x – y)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(x – y + 7) (y – x + 7)

**26. a ^{2} + 4b^{2} – 4ab – 4c^{2}**

**Solution:**

We have,

a^{2} + 4b^{2} – 4ab – 4c^{2}

a^{2} – 2 (a) (2b) + (2b)^{2} – (2c)^{2}

(a – 2b)^{2} – (2c)^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a – 2b + 2c) (a – 2b – 2c)

**27. x ^{2} – y^{2} – 4xz + 4z^{2}**

**Solution:**

We have,

x^{2} – y^{2} – 4xz + 4z^{2}

x^{2} – 2 (x) (2z) + (2z)^{2} – y^{2}

As (a-b)^{2} = a^{2} + b^{2} – 2ab

(x – 2z)^{2} – y^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(x + y – 2z) (x – y – 2z)

### EXERCISE 7.7 PAGE NO: 7.27

**Factorize each of the following algebraic expressions:**

**1. x ^{2} + 12x – 45**

**Solution:**

We have,

x^{2} + 12x – 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 with 15x – 3x

-45 by 15 × 3

x^{2} + 12x – 45 = x^{2} + 15x – 3x – 45

= x (x + 15) – 3 (x + 15)

= (x – 3) (x + 15)

**2. 40 + 3x – x ^{2}**

**Solution:**

We have,

40 + 3x – x^{2}

-(x^{2} – 3x – 40)

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

So we can replace -3x with 5x – 8x

-40 by 5 × -8

-(x^{2} – 3x – 40) = x^{2} + 5x – 8x – 40

= -x (x + 5) – 8 (x + 5)

= -(x – 8) (x + 5)

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

**3. a ^{2} + 3a – 88**

**Solution:**

We have,

a^{2} + 3a – 88

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

So we can replace 3a with 11a – 8a

-40 by -11 × 8

a^{2} + 3a – 88 = a^{2} + 11a – 8a – 88

= a (a + 11) – 8 (a + 11)

= (a – 8) (a + 11)

**4. a ^{2} – 14a – 51**

**Solution:**

We have,

a^{2} – 14a – 51

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

So we can replace -14a with 3a – 17a

-51 by -17 × 3

a^{2} – 14a – 51 = a^{2} + 3a – 17a – 51

= a (a + 3) – 17 (a + 3)

= (a – 17) (a + 3)

**5. x ^{2} + 14x + 45**

**Solution:**

We have,

x^{2} + 14x + 45

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

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

45 by 5 × 9

x^{2} + 14x + 45 = x^{2} + 5x + 9x + 45

= x (x + 5) – 9 (x + 5)

= (x + 9) (x + 5)

**6. x ^{2} – 22x + 120**

**Solution:**

We have,

x^{2} – 22x + 120

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

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

120 by -12 × -10

x^{2} – 22x + 120 = x^{2} – 12x – 10x + 120

= x (x – 12) – 10 (x – 12)

= (x – 10) (x – 12)

**7. x ^{2} – 11x – 42**

**Solution:**

We have,

x^{2} – 11x – 42

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

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

-42 by 3 × -14

x^{2} – 11x – 42 = x^{2} + 3x – 14x – 42

= x (x + 3) – 14 (x + 3)

= (x – 14) (x + 3)

**8. a ^{2} + 2a – 3**

**Solution:**

We have,

a^{2} + 2a – 3

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

So we can replace 2a with 3a -a

-3 by 3 × -1

a^{2} + 2a – 3 = a^{2} + 3a – a – 3

= a (a + 3) – 1 (a + 3)

= (a – 1) (a + 3)

**9. a ^{2} + 14a + 48**

**Solution:**

We have,

a^{2} + 14a + 48

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

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

48 by 8 × 6

a^{2} + 14a + 48 = a^{2} + 8a + 6a + 48

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

= (a + 6) (a + 8)

**10. x ^{2} – 4x – 21**

**Solution:**

We have,

x^{2} – 4x – 21

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

So we can replace -4x with 3x – 7x

-21 by 3 × -7

x^{2} + 4x – 21 = x^{2} + 3x – 7x – 21

= x (x + 3) – 7 (x + 3)

= (x – 7) (x + 3)

**11. y ^{2} + 5y – 36**

**Solution:**

We have,

y^{2} + 5y – 36

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

So we can replace 5y with 9y – 4y

-36 by 9 × -4

y^{2} + 5y – 36 = y^{2} + 9y – 4y – 36

= y (y + 9) – 4 (y + 9)

= (y – 4) (y + 9)

**12. (a ^{2} – 5a)^{2} – 36**

**Solution:**

We have,

(a^{2} – 5a)^{2} – 36

(a^{2} – 5a)^{2} – 6^{2}

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

(a^{2} – 5a)^{2} – 6^{2} = (a^{2} – 5a + 6) (a^{2} – 5a – 6)

So now we shall factorize the expression (a^{2} – 5a + 6)

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

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

6 by 1 × -6

a^{2} -5a – 6 = a^{2} + a – 6a – 6

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

= (a – 6) (a + 1)

So now we shall factorize the expression (a^{2} – 5a + 6)

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

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

6 by -2 × -3

a^{2} -5a + 6 = a^{2} – 2a – 3a + 6

= a (a – 2) -3 (a – 2)

= (a – 3) (a – 2)

∴ (a^{2} – 5a)^{2} – 36 = (a^{2} – 5a + 6) (a^{2} – 5a – 6)

= (a + 1) (a – 6) (a – 2) (a – 3)

**13. (a + 7) (a – 10) + 16**

**Solution:**

We have,

(a + 7) (a – 10) + 16

a^{2} – 10a + 7a – 70 + 16

a^{2} – 3a – 54

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

So we can replace -3a with 6a – 9a

-54 by 6 × -9

a^{2} – 3a – 54 = a^{2} + 6a – 9a – 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. 2x ^{2} + 5x + 3**

**Solution:**

We have,

2x^{2} + 5x + 3

The coefficient of x^{2} is 2

The coefficient of x is 5

Constant term is 3

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

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

2x^{2} + 5x + 3 = 2x^{2} + 2x + 3x + 3

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

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

**2. 2x ^{2} – 3x – 2**

**Solution:**

We have,

2x^{2} – 3x – 2

The coefficient of x^{2} is 2

The coefficient of x is -3

Constant term is -2

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

2x^{2} – 3x – 2 = 2x^{2} – 4x + x – 2

= 2x (x – 2) + 1 (x – 2)

= (x – 2) (2x + 1)

**3. 3x ^{2} + 10x + 3**

**Solution:**

We have,

3x^{2} + 10x + 3

The coefficient of x^{2} is 3

The coefficient of x is 10

Constant term is 3

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

3x^{2} + 10x + 3 = 3x^{2} + 9x + x + 3

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

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

**4. 7x – 6 – 2x ^{2}**

**Solution:**

We have,

7x – 6 – 2x^{2}

– 2x^{2} + 7x – 6

2x^{2} – 7x + 6

The coefficient of x^{2} is 2

The coefficient of x is -7

Constant term is 6

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

2x^{2} – 7x + 6 = 2x^{2} – 4x – 3x + 6

= 2x (x – 2) – 3 (x – 2)

= (x – 2) (2x – 3)

**5. 7x ^{2} – 19x – 6**

**Solution:**

We have,

7x^{2} – 19x – 6

The coefficient of x^{2} is 7

The coefficient of x is -19

Constant term is -6

So, we express the middle term -19x as 2x – 21x

7x^{2} – 19x – 6 = 7x^{2} + 2x – 21x – 6

= x (7x + 2) – 3 (7x + 2)

= (7x + 2) (x – 3)

**6. 28 – 31x – 5x ^{2}**

**Solution:**

We have,

28 – 31x – 5x^{2}

– 5x^{2} -31x + 28

5x^{2} + 31x – 28

The coefficient of x^{2} is 5

The coefficient of x is 31

Constant term is -28

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

5x^{2} + 31x – 28 = 5x^{2} – 4x + 35x – 28

= x (5x – 4) + 7 (5x – 4)

= (x + 7) (5x – 4)

**7. 3 + 23y – 8y ^{2}**

**Solution:**

We have,

3 + 23y – 8y^{2}

– 8y^{2} + 23y + 3

8y^{2} – 23y – 3

The coefficient of y^{2} is 8

The coefficient of y is -23

Constant term is -3

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

8y^{2} – 23y – 3 = 8y^{2} – 24y + y – 3

= 8y (y – 3) + 1 (y – 3)

= (8y + 1) (y – 3)

**8. 11x ^{2} – 54x + 63**

**Solution:**

We have,

11x^{2} – 54x + 63

The coefficient of x^{2} is 11

The coefficient of x is -54

Constant term is 63

So, we express the middle term -54x as -33x – 21x

11x^{2} – 54x + 63 = 11x^{2} – 33x – 21x + 63

= 11x (x – 3) – 21 (x – 3)

= (11x – 21) (x – 3)

**9. 7x – 6x ^{2} + 20**

**Solution:**

We have,

7x – 6x^{2} + 20

– 6x^{2} + 7x + 20

6x^{2} – 7x – 20

The coefficient of x^{2} is 6

The coefficient of x is -7

Constant term is -20

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

6x^{2} – 7x – 20 = 6x^{2} – 15x + 8x – 20

= 3x (2x – 5) + 4 (2x – 5)

= (3x + 4) (2x – 5)

**10. 3x ^{2} + 22x + 35**

**Solution:**

We have,

3x^{2} + 22x + 35

The coefficient of x^{2} is 3

The coefficient of x is 22

Constant term is 35

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

3x^{2} + 22x + 35 = 3x^{2} + 15x + 7x + 35

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

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

**11. 12x ^{2} – 17xy + 6y^{2}**

**Solution:**

We have,

12x^{2} – 17xy + 6y^{2}

The coefficient of x^{2} is 12

The coefficient of x is -17y

Constant term is 6y^{2}

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

12x^{2} -17xy+ 6y^{2} = 12x^{2} – 9xy – 8xy + 6y^{2}

= 3x (4x – 3y) – 2y (4x – 3y)

= (3x – 2y) (4x – 3y)

**12. 6x ^{2} – 5xy – 6y^{2}**

**Solution:**

We have,

6x^{2} – 5xy – 6y^{2}

The coefficient of x^{2} is 6

The coefficient of x is -5y

Constant term is -6y^{2}

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

6x^{2} -5xy- 6y^{2} = 6x^{2} + 4xy – 9xy – 6y^{2}

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

= (2x – 3y) (3x + 2y)

**13. 6x ^{2} – 13xy + 2y^{2}**

**Solution:**

We have,

6x^{2} – 13xy + 2y^{2}

The coefficient of x^{2} is 6

The coefficient of x is -13y

Constant term is 2y^{2}

So, we express the middle term -13xy as -12xy – xy

6x^{2} -13xy+ 2y^{2} = 6x^{2} – 12xy – xy + 2y^{2}

= 6x (x – 2y) – y (x – 2y)

= (6x – y) (x – 2y)

**14. 14x ^{2} + 11xy – 15y^{2}**

**Solution:**

We have,

14x^{2} + 11xy – 15y^{2}

The coefficient of x^{2} is 14

The coefficient of x is 11y

Constant term is -15y^{2}

So, we express the middle term 11xy as 21xy – 10xy

14x^{2} + 11xy- 15y^{2} = 14x^{2} + 21xy – 10xy – 15y^{2}

= 2x (7x – 5y) + 3y (7x – 5y)

= (2x + 3y) (7x – 5y)

**15. 6a ^{2} + 17ab – 3b^{2}**

**Solution:**

We have,

6a^{2} + 17ab – 3b^{2}

The coefficient of a^{2} is 6

The coefficient of a is 17b

Constant term is -3b^{2}

So, we express the middle term 17ab as 18ab – ab

6a^{2} +17ab– 3b^{2} = 6a^{2} + 18ab – ab – 3b^{2}

= 6a (a + 3b) – b (a + 3b)

= (6a – b) (a + 3b)

**16. 36a ^{2} + 12abc – 15b^{2}c^{2}**

**Solution:**

We have,

36a^{2} + 12abc – 15b^{2}c^{2}

The coefficient of a^{2} is 36

The coefficient of a is 12bc

Constant term is -15b^{2}c^{2}

So, we express the middle term 12abc as 30abc – 18abc

36a^{2} –12abc– 15b^{2}c^{2} = 36a^{2} + 30abc – 18abc – 15b^{2}c^{2}

= 6a (6a + 5bc) – 3bc (6a + 5bc)

= (6a + 5bc) (6a – 3bc)

= (6a + 5bc) 3(2a – bc)

**17. 15x ^{2} – 16xyz – 15y^{2}z^{2}**

**Solution:**

We have,

15x^{2} – 16xyz – 15y^{2}z^{2}

The coefficient of x^{2} is 15

The coefficient of x is -16yz

Constant term is -15y^{2}z^{2}

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

15x^{2} -16xyz- 15y^{2}z^{2} = 15x^{2} – 25yz + 9yz – 15y^{2}z^{2}

= 5x (3x – 5yz) + 3yz (3x – 5yz)

= (5x + 3yz) (3x – 5yz)

**18. (x – 2y) ^{2} – 5 (x – 2y) + 6**

**Solution:**

We have,

(x – 2y)^{2} – 5 (x – 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 – 2y) as -2(x – 2y) -3(x – 2y)

(x – 2y)^{2} – 5 (x – 2y) + 6 = (x – 2y)^{2} – 2 (x – 2y) – 3 (x – 2y) + 6

= (x – 2y – 2) (x – 2y – 3)

**19. (2a – b) ^{2} + 2 (2a – b) – 8**

**Solution:**

We have,

(2a – b)^{2} + 2 (2a – b) – 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 – b) as 4 (2a –b) – 2 (2a – b)

(2a – b)^{2} + 2 (2a – b) – 8 = (2a – b)^{2} + 4 (2a – b) – 2 (2a – b) – 8

= (2a – b) (2a – b + 4) – 2 (2a – b + 4)

= (2a – b + 4) (2a – b – 2)

### EXERCISE 7.9 PAGE NO: 7.32

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

**1. p ^{2} + 6p + 8**

**Solution:**

We have,

p^{2} + 6p + 8

The coefficient of p^{2} is unity. So, we add and subtract the square of half of the coefficient of p.

p^{2} + 6p + 8 = p^{2} + 6p + 3^{2} – 3^{2} + 8 (Adding and subtracting 3^{2})

= (p + 3)^{2} – 1^{2} (By completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (p + 3 – 1) (p + 3 + 1)

= (p + 2) (p + 4)

**2. q ^{2} – 10q + 21**

**Solution:**

We have,

q^{2} – 10q + 21

The coefficient of q^{2} is unity. So, we add and subtract the square of half of the coefficient of q.

q^{2} – 10q + 21 = q^{2} – 10q+ 5^{2} – 5^{2} + 21 (Adding and subtracting 5^{2})

= (q – 5)^{2} – 2^{2} (By completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (q – 5 – 2) (q – 5 + 2)

= (q – 3) (q – 7)

**3. 4y ^{2} + 12y + 5**

**Solution:**

We have,

4y^{2} + 12y + 5

4(y^{2} + 3y + 5/4)

The coefficient of y^{2} is unity. So, we add and subtract the square of half of the coefficient of y.

4(y^{2} + 3y + 5/4) = 4 [y^{2} + 3y + (3/2)^{2} – (3/2)^{2} + 5/4] (Adding and subtracting (3/2)^{2})

= 4 [(y + 3/2)^{2} – 1^{2}] (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= 4 (y + 3/2 + 1) (y + 3/2 – 1)

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

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

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

**4. p ^{2} + 6p – 16**

**Solution:**

We have,

p^{2} + 6p – 16

The coefficient of p^{2} is unity. So, we add and subtract the square of half of the coefficient of p.

p^{2} + 6p – 16 = p^{2} + 6p + 3^{2} – 3^{2} – 16 (Adding and subtracting 3^{2})

= (p + 3)^{2} – 5^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (p + 3 + 5) (p + 3 – 5)

= (p + 8) (p – 2)

**5. x ^{2} + 12x + 20**

**Solution:**

We have,

x^{2} + 12x + 20

The coefficient of x^{2} is unity. So, we add and subtract the square of half of the coefficient of x.

x^{2} + 12x + 20 = x^{2} + 12x + 6^{2} – 6^{2} + 20 (Adding and subtracting 6^{2})

= (x + 6)^{2} – 4^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (x + 6 + 4) (x + 6 – 4)

= (x + 2) (x + 10)

**6. a ^{2} – 14a – 51**

**Solution:**

We have,

a^{2} – 14a – 51

The coefficient of a^{2} is unity. So, we add and subtract the square of half of the coefficient of a.

a^{2} – 14a – 51 = a^{2} – 14a + 7^{2} – 7^{2} – 51 (Adding and subtracting 7^{2})

= (a – 7)^{2} – 10^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (a – 7 + 10) (9 – 7 – 10)

= (a – 17) (a + 3)

**7. a ^{2} + 2a – 3**

**Solution:**

We have,

a^{2} + 2a – 3

The coefficient of a^{2} is unity. So, we add and subtract the square of half of the coefficient of a.

a^{2} + 2a – 3 = a^{2} + 2a + 1^{2} – 1^{2} – 3 (Adding and subtracting 1^{2})

= (a + 1)^{2} – 2^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (a + 1 + 2) (a + 1 – 2)

= (a + 3) (a – 1)

**8. 4x ^{2} – 12x + 5**

**Solution:**

We have,

4x^{2} – 12x + 5

4(x^{2} – 3x + 5/4)

The coefficient of x^{2} is unity. So, we add and subtract the square of half of the coefficient of x.

4(x^{2} – 3x + 5/4) = 4 [x^{2} – 3x + (3/2)^{2} – (3/2)^{2} + 5/4] (Adding and subtracting (3/2)^{2})

= 4 [(x – 3/2)^{2} – 1^{2}] (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= 4 (x – 3/2 + 1) (x – 3/2 – 1)

= 4 (x – 1/2) (x – 5/2) (by taking LCM)

= 4 [(2x-1)/2] [(2x – 5)/2]

= (2x – 5) (2x – 1)

**9. y ^{2} – 7y + 12**

**Solution:**

We have,

y^{2} – 7y + 12

The coefficient of y^{2} is unity. So, we add and subtract the square of half of the coefficient of y.

y^{2} – 7y + 12 = y^{2} – 7y + (7/2)^{2} – (7/2)^{2} + 12 [Adding and subtracting (7/2)^{2}]

= (y – 7/2)^{2} – (7/2)^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (y – (7/2- 1/2)) (y – (7/2 + 1/2))

= (y – 3) (y – 4)

**10. z ^{2} – 4z – 12**

**Solution:**

We have,

z^{2} – 4z – 12

The coefficient of z^{2} is unity. So, we add and subtract the square of half of the coefficient of z.

z^{2} – 4z – 12 = z^{2} – 4z + 2^{2} – 2^{2} – 12 [Adding and subtracting 2^{2}]

= (z – 2)^{2} – 4^{2} (Completing the square)

By using the formula (a^{2} – b^{2}) = (a+b) (a-b)

= (z – 2 + 4) (z – 2 – 4)

= (z – 6) (z + 2)

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