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 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

Download PDF Here

carouselExampleControls111

Previous



Next

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. 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 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 x²

∴ The greatest common factor = 2x²

2. 6x³y and 18x²y³

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 x²y

∴ 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

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. 42x²yz and 63x³y²z³

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 x²yz

∴ The greatest common factor = 21x²yz

5. 12ax², 6a²x³ and 2a³x⁵

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 ax²

∴ 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

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. 4a²b³, -12a³b, 18a⁴b³

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 a²b

∴ The greatest common factor = 2a²b

8. 6x²y², 9xy³, 3x³y²

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 xy²

∴ The greatest common factor = 3xy²

9. a²b³, a³b²

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 a²b²

∴ 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 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 a²c²

∴ The greatest common factor = 18a²c²

11. x³, -yx²

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 x²

∴ The greatest common factor = x²

12. 15a³, -45a², -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. 2x³y², 10x²y³, 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. 14x³y⁵, 10x⁵y³, 2x²y²

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 x²y²

∴ 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 – 9

Solution:

The greatest common factor in the given two terms is 3

3x – 9

3 (x – 3)

2. 5x – 15x²

Solution:

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

5x – 15x²

5x (1 – 3x)

3. 20a¹²b² – 15a⁸b⁴

Solution:

Greatest common factor in the given two terms is 5a⁸b²

20a¹²b² – 15a⁸b⁴

5a⁸b² (4a⁴ – 3b²)

4. 72x⁶y⁷ – 96x⁷y⁶
Solution:

Greatest common factor in the given two terms is 24x⁶y⁶

72x⁶y⁷ – 96x⁷y⁶

24x⁶y⁶ (3y – 4x)

5. 20x³ – 40x² + 80x

Solution:

Greatest common factor in the given three terms is 20x

20x³ – 40x² + 80x

20x (x² – 2x +4)

6. 2x³y² – 4x²y³ + 8xy⁴

Solution:

Greatest common factor in the given three terms is 2xy²

2x³y² – 4x²y³ + 8xy⁴

2xy² (x² – 2xy + 4y²)

7. 10m³n² + 15m⁴n – 20m²n³

Solution:

Greatest common factor in the given three terms is 5mn²

10m³n² + 15m⁴n – 20m²n³

5m²n (2mn + 3m² – 4n²)

8. 2a⁴b⁴ – 3a³b⁵ + 4a²b⁵

Solution:

Greatest common factor in the given three terms is a²b⁴

2a⁴b⁴ – 3a³b⁵ + 4a²b⁵

a²b⁴ (2a² – 3ab + 4b)

9. 28a² + 14a²b² – 21a⁴

Solution:

Greatest common factor in the given three terms is 7a²

28a² + 14a²b² – 21a⁴

7a² (4a + 2b² – 3a²)

10. a⁴b – 3a²b² – 6ab³

Solution:

Greatest common factor in the given three terms is ab

a⁴b – 3a²b² – 6ab³

ab (a³ – 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. x⁴y² – x²y⁴ – x⁴y⁴

Solution:

Greatest common factor in the given three terms is x²y²

x⁴y² – x²y⁴ – x⁴y⁴

x²y² (x² – y² – x²y²)

13. 9x²y + 3axy

Solution:

Greatest common factor in the given three terms is 3xy

9x²y + 3axy

3xy (3x + 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 – 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) – 12a² (6a – 5b)

Solution:

We have,

9a (6a – 5b) – 12a² (6a – 5b)

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

(9a – 12a²) (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. 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,

(a² + b² + 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 – 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 – 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,

(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,

4(x – 2y) (-x + 2y + 2)

13. x³ (a – 2b) + x² (a – 2b)

Solution:

We have,

x³ (a – 2b) + x² (a – 2b)

By taking x² (a – 2b) as common, we get,

(x + 1) [x² (a – 2b)]

x² (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. p²q – pr² – pq + r²

Solution:

We have,

p²q – pr² – pq + r²

By grouping similar terms, we get,

p²q – pq – pr² + r²

pq(p – 1) –r²(p – 1)

(p – 1) (pq – 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 – bx – by

Solution:

We have,

ax + ay – bx – by

a(x + y) –b (x + y)

(a – b) (x + y)

5. xa² + xb² – ya² – yb²

Solution:

We have,

xa² + xb² – ya² – yb²

x(a² + b²) –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 – by – ay + y²

Solution:

We have,

ab – by – ay + y²

By grouping similar terms, we get,

Ab – ay – by + y²

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. lm² – mn² – lm + n²

Solution:

We have,

lm² – mn² – lm + n²

By grouping similar terms, we get,

lm² – lm – mn² + n²

lm (m – 1) – n² (m – 1)

(lm – n²) (m – 1)

11. x³ – y² + x – x²y²

Solution:

We have,

x³ – y² + x – x²y²

By grouping similar terms, we get,

x + x³ – y² – x²y²

x (1 + x²) – y² (1 + x²)

(x – y²) (1 + x²)

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. x² – 2ax – 2ab + bx

Solution:

We have,

x² – 2ax – 2ab + bx

By grouping similar terms, we get,

x² + bx – 2ax – 2ab

x (x + b) – 2a (x + b)

(x – 2a) (x + b)

14. x³ – 2x²y + 3xy² – 6y³

Solution:

We have,

x³ – 2x²y + 3xy² – 6y³

By grouping similar terms, we get,

x³ + 3xy² – 2x²y – 6y³

x (x² + 3y²) – 2y (x² + 3y²)

(x – 2y) (x² + 3y²)

15. abx² + (ay – b) x – y

Solution:

We have,

abx² + (ay – b) x – y

abx² + ayx – bx – y

By grouping similar terms, we get,

abx² – 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

a²x² + b²y² + 2axby + b²x² + a²y² – 2axby

a²x² + b²y² + b²x² + a²y²

By grouping similar terms, we get,

a²x² + a²y² + b²y² + b²x²

a² (x² + y²) + b² (x² + y²)

(a² + b²) (x² + y²)

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 (x² + 1) + x (a² + b²)

Solution:

We have,

ab(x² + 1) + x(a² + b²)

abx² + ab + xa² + xb²

By grouping similar terms, we get,

abx² + 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

a²x² + ax³ + x + a

ax² (a + x) + 1 (x + a)

(x + a) (ax² + 1)

20. a (a – 2b – c) + 2bc

Solution:

We have,

a (a – 2b – c) + 2bc

a² – 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

a² + ab – ac – bc

a (a + b) – c (a + b)

(a + b) (a – c)

22. x² – 11xy – x + 11y

Solution:

We have,

x² – 11xy – x + 11y

By grouping similar terms, we get,

x² – 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. x² + y – xy – x

Solution:

We have,

x² + y – xy – x

By grouping similar terms, we get,

x² – 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. 16x² – 25y²

Solution:

We have,

16x² – 25y²

(4x)² – (5y)²

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

(4x + 5y) (4x – 5y)

2. 27x² – 12y²

Solution:

We have,

27x² – 12y²

By taking 3 as common, we get,

3 [(3x)² – (2y)²]

By using the formula (a² – b²) = (a-b) (a+b)

3 (3x + 2y) (3x – 2y)

3. 144a² – 289b²

Solution:

We have,

144a² – 289b²

(12a)² – (17b)²

By using the formula (a² – b²) = (a-b) (a+b)

(12a + 17b) (12a – 17b)

4. 12m² – 27

Solution:

We have,

12m² – 27

By taking 3 as common, we get,

3 (4m² – 9)

3 [(2m)² – 3²]

By using the formula (a² – b²) = (a-b) (a+b)

3 (2m + 3) (2m – 3)

5. 125x² – 45y²

Solution:

We have,

125x² – 45y²

By taking 5 as common, we get,

5 (25x² – 9y²)

5 [(5x)² – (3y)²]

By using the formula (a² – b²) = (a-b) (a+b)

5 (5x + 3y) (5x – 3y)

6. 144a² – 169b²

Solution:

We have,

144a² – 169b²

(12a)² – (13b)²

By using the formula (a² – b²) = (a-b) (a+b)

(12a + 13b) (12a – 13b)

7. (2a – b)² – 16c²

Solution:

We have,

(2a – b)² – 16c²

(2a – b)² – (4c)²

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

(x + 4x + 2y – 2y) (x – 4x + 2y + 2y)

(5x) (4y – 3x)

9. 3a⁵ – 48a³

Solution:

We have,

3a⁵ – 48a³

By taking 3 as common, we get,

3a³ (a² – 16)

3a³ (a² – 4²)

By using the formula (a² – b²) = (a-b) (a+b)

3a³ (a + 4) (a – 4)

10. a⁴ – 16b⁴

Solution:

We have,

a⁴ – 16b⁴

(a²)² – (4b²)²

By using the formula (a² – b²) = (a-b) (a+b)

(a² + 4b²) (a² – 4b²)

By using the formula (a² – b²) = (a-b) (a+b)

(a² + 4b²) (a + 2b) (a – 2b)

11. x⁸ – 1

Solution:

We have,

x⁸ – 1

(x⁴)²–(1)²

By using the formula (a² – b²) = (a-b) (a+b)

(x⁴ + 1) (x⁴ – 1)

By using the formula (a² – b²) = (a-b) (a+b)

(x⁴ + 1) (x² + 1) (x – 1) (x + 1)

12. 64 – (a + 1)²

Solution:

We have,

64 – (a + 1)²

8² – (a + 1)²

By using the formula (a² – b²) = (a-b) (a+b)

(a + 9) (7 – a)

13. 36l² – (m + n)²

Solution:

We have,

36l² – (m + n)²

(6l)² – (m + n)²

By using the formula (a² – b²) = (a-b) (a+b)

(6l + m + n) (6l – m – n)

14. 25x⁴y⁴ – 1

Solution:

We have,

25x⁴y⁴ – 1

(5x²y²)² – (1)²

By using the formula (a² – b²) = (a-b) (a+b)

(5x²y² – 1) (5x²y² + 1)

15. a⁴ – 1/b⁴

Solution:

We have,

a⁴ – 1/b⁴

(a²)² – (1/b²)²

By using the formula (a² – b²) = (a-b) (a+b)

(a² + 1/b²) (a² – 1/b²)

By using the formula (a² – b²) = (a-b) (a+b)

(a² + 1/b²) (a – 1/b) (a + 1/b)

16. x³ – 144x

Solution:

We have,

x³ – 144x

x [x² – (12)²]

By using the formula (a² – b²) = (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 (a² – b²) = (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

By using the formula (a² – b²) = (a-b) (a+b)

19. (3 + 2a)² – 25a²

Solution:

We have,

(3 + 2a)² – 25a²

(3 + 2a)² – (5a)²

By using the formula (a² – b²) = (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 (a² – b²) = (a-b) (a+b)

(x + y + a – b) (x + y – a + b)

21. 1/16x²y² – 4/49y²z²

Solution:

We have,

1/16x²y² – 4/49y²z²

(1/4xy)² – (2/7yz)²

By using the formula (a² – b²) = (a-b) (a+b)

(xy/4 + 2yz/7) (xy/4 – 2yz/7)

y² (x/4 + 2/7z) (x/4 – 2/7z)

22. 75a³b² – 108ab⁴

Solution:

We have,

75a³b² – 108ab⁴

3ab² (25a² – 36b²)

3ab² [(5a)² – (6b)²]

By using the formula (a² – b²) = (a-b) (a+b)

3ab² (5a + 6b) (5a – 6b)

23. x⁵ – 16x³

Solution:

We have,

x⁵ – 16x³

x³ (x² – 16)

x³ (x² – 4²)

By using the formula (a² – b²) = (a-b) (a+b)

x³ (x + 4) (x – 4)

24. 50/x² – 2x²/81

Solution:

We have,

50/x² – 2x²/81

2 (25/x² – x²/81)

2 [(5/x)² – (x/9)²]

By using the formula (a² – b²) = (a-b) (a+b)

2 (5/x+ x/9) (5/x – x/9)

25. 256x³ – 81x

Solution:

We have,

256x³ – 81x

x (256x⁴ – 81)

x [(16x²)² – 9²]

By using the formula (a² – b²) = (a-b) (a+b)

x (4x + 3) (4x – 3) (16x² + 9)

26. a⁴ – (2b + c)⁴

Solution:

We have,

a⁴ – (2b + c)⁴

(a²)² – [(2b + c)²]²

By using the formula (a² – b²) = (a-b) (a+b)

By using the formula (a² – b²) = (a-b) (a+b)

27. (3x + 4y)⁴ – x⁴

Solution:

We have,

(3x + 4y)⁴ – x⁴

By using the formula (a² – b²) = (a-b) (a+b)

28. p²q² – p⁴q⁴

Solution:

We have,

p²q² – p⁴q⁴

(pq)² – (p²q²)²

By using the formula (a² – b²) = (a-b) (a+b)

(pq + p²q²) (pq – p²q²)

p²q² (1 + pq) (1 – pq)

29. 3x³y – 243xy³

Solution:

We have,

3x³y – 243xy³

3xy (x² – 81y²)

3xy [x² – (9y)²]

By using the formula (a² – b²) = (a-b) (a+b)

(3xy) (x + 9y) (x – 9y)

30. a⁴b⁴ – 16c⁴

Solution:

We have,

a⁴b⁴ – 16c⁴

(a²b²)² – (4c²)²

By using the formula (a² – b²) = (a-b) (a+b)

(a²b² + 4c²) (a²b² – 4c²)

By using the formula (a² – b²) = (a-b) (a+b)

(a²b² + 4c²) (ab + 2c) (ab – 2c)

31. x⁴ – 625

Solution:

We have,

x⁴ – 625

(x²)² – (25)²

By using the formula (a² – b²) = (a-b) (a+b)

(x² + 25) (x² – 25)

(x² + 25) (x² – 5²)

By using the formula (a² – b²) = (a-b) (a+b)

(x² + 25) (x + 5) (x – 5)

32. x⁴ – 1

Solution:

We have,

x⁴ – 1

(x²)² – (1)²

By using the formula (a² – b²) = (a-b) (a+b)

(x² + 1) (x² – 1)

By using the formula (a² – b²) = (a-b) (a+b)

(x² + 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

By using the formula (a² – b²) = (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 – x² + y²

Solution:

We have,

x – y – x² + y²

x – y – (x² – y²)

By using the formula (a² – b²) = (a-b) (a+b)

x – y – (x + y) (x – y)

(x – y) (1 – x – y)

35. 16(2x – 1)² – 25y²

Solution:

We have,

16(2x – 1)² – 25y²

By using the formula (a² – b²) = (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

By using the formula (a² – b²) = (a-b) (a+b)

(2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)

(2xy + 3x – 1) (2xy – 3x + 5)

37. (2x + 1)² – 9x⁴

Solution:

We have,

(2x + 1)² – 9x⁴

(2x + 1)² – (3x²)²

By using the formula (a² – b²) = (a-b) (a+b)

(2x + 1 + 3x²) (2x + 1 – 3x²)

(3x² + 2x + 1) (-3x² + 2x + 1)

38. x⁴ – (2y – 3z)²

Solution:

We have,

x⁴ – (2y – 3z)²

(x²)² – (2y – 3z)²

By using the formula (a² – b²) = (a-b) (a+b)

(x² + 2y – 3z) (x² – 2y + 3z)

39. a² – b² + a – b

Solution:

We have,

a² – b² + a – b

By using the formula (a² – b²) = (a-b) (a+b)

(a + b) (a – b) + (a – b)

(a – b) (a + b + 1)

40. 16a⁴ – b⁴

Solution:

We have,

16a⁴ – b⁴

(4a²)² – (b²)²

(4a² + b²) (4a² – b²)

By using the formula (a² – b²) = (a-b) (a+b)

(4a² + b²) (2a + b) (2a – b)

41. a⁴ – 16(b – c)⁴

Solution:

We have,

a⁴ – 16(b – c)⁴

(a²)² – [4 (b – c)²]

By using the formula (a² – b²) = (a-b) (a+b)

By using the formula (a² – b²) = (a-b) (a+b)

42. 2a⁵ – 32a

Solution:

We have,

2a⁵ – 32a

2a (a⁴ – 16)

2a [(a²)² – (4)²]

By using the formula (a² – b²) = (a-b) (a+b)

2a (a² + 4) (a² – 4)

2a (a² + 4) (a² – 2²)

By using the formula (a² – b²) = (a-b) (a+b)

2a (a² + 4) (a + 2) (a – 2)

43. a⁴b⁴ – 81c⁴

Solution:

We have,

a⁴b⁴ – 81c⁴

(a²b²)² – (9c²)²

By using the formula (a² – b²) = (a-b) (a+b)

(a²b² + 9c²) (a²b² – 9c²)

By using the formula (a² – b²) = (a-b) (a+b)

(a²b² + 9c²) (ab + 3c) (ab – 3c)

44. xy⁹ – yx⁹

Solution:

We have,

xy⁹ – yx⁹

-xy (x⁸ – y⁸)

-xy [(x⁴)² – (y⁴)²]

By using the formula (a² – b²) = (a-b) (a+b)

-xy (x⁴ + y⁴) (x⁴ – y⁴)

By using the formula (a² – b²) = (a-b) (a+b)

-xy (x⁴ + y⁴) (x² + y²) (x² – y²)

By using the formula (a² – b²) = (a-b) (a+b)

-xy (x⁴ + y⁴) (x² + y²) (x + y) (x – y)

45. x³ – x

Solution:

We have,

x³ – x

x (x² – 1)

By using the formula (a² – b²) = (a-b) (a+b)

x (x + 1) (x – 1)

46. 18a²x² – 32

Solution:

We have,

18a²x² – 32

2 [(3ax)² – (4)²]

By using the formula (a² – b²) = (a-b) (a+b)

2 (3ax + 4) (3ax – 4)

EXERCISE 7.6 PAGE NO: 7.22

Factorize each of the following algebraic expressions:

1. 4x² + 12xy + 9y²

Solution:

We have,

4x² + 12xy + 9y²

By using the formula (x + y)² = x² + y² + 2xy

(2x)² + (3y)² + 2 (2x) (3y)

(2x + 3y)²

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

2. 9a² – 24ab + 16b²

Solution:

We have,

9a² – 24ab + 16b²

By using the formula (x – y)² = x² + y² – 2xy

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

(3a)² + (4b)² – 2 (3a) (4b)

(3a – 4b)²

(3a – 4b) (3a – 4b)

3. p²q² – 6pqr + 9r²

Solution:

We have,

p²q² – 6pqr + 9r²

By using the formula (a – b)² = a² + b² – 2ab

(pq)² + (3r)² – 2 (pq) (3r)

(pq – 3r)²

(pq – 3r) (pq – 3r)

4. 36a² + 36a + 9

Solution:

We have,

36a² + 36a + 9

(6a)² + 2 × 6a × 3 + 3²

(6a + 3)²

5. a² + 2ab + b² – 16

Solution:

We have,

a² + 2ab + b² – 16

By using the formula (a – b)² = a² + b² – 2ab

(a + b)² – 4²

By using the formula (a² – b²) = (a+b) (a-b)

(a + b + 4) (a + b – 4)

6. 9z² – x² + 4xy – 4y²

Solution:

We have,

9z² – x² + 4xy – 4y²

(3z)² – [x² – 2 (x) (2y) + (2y)²]

By using the formula (a – b)² = a² + b² – 2ab

(3z)² – (x – 2y)²

By using the formula (a² – b²) = (a+b) (a-b)

7. 9a⁴ – 24a²b² + 16b⁴ – 256

Solution:

We have,

9a⁴ – 24a²b² + 16b⁴ – 256

(3a²)² – 2 (4a²) (3b²) + (4b²)² – (16)²

By using the formula (a – b)² = a² + b² – 2ab

(3a² – 4b²)² – (16)²

By using the formula (a² – b²) = (a+b) (a-b)

(3a² – 4b² + 16) (3a² – 4b² – 16)

8. 16 – a⁶ + 4a³b³ – 4b⁶

Solution:

We have,

16 – a⁶ + 4a³b³ – 4b⁶

4² – [(a³)² – 2 (a³) (2b³) + (2b³)²]

By using the formula (a – b)² = a² + b² – 2ab

4² – (a³ – 2b³)²

By using the formula (a² – b²) = (a+b) (a-b)

9. a² – 2ab + b² – c²

Solution:

We have,

a² – 2ab + b² – c²

By using the formula (a – b)² = a² + b² – 2ab

(a – b)² – c²

By using the formula (a² – b²) = (a+b) (a-b)

(a – b + c) (a – b – c)

10. x² + 2x + 1 – 9y²

Solution:

We have,

x² + 2x + 1 – 9y²

By using the formula (a – b)² = a² + b² – 2ab

(x + 1)² – (3y)²

By using the formula (a² – b²) = (a+b) (a-b)

(x + 3y + 1) (x – 3y + 1)

11. a² + 4ab + 3b²

Solution:

We have,

a² + 4ab + 3b²

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

a² + ab + 3ab + 3b²

By grouping we get,

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

(a + 3b) (a + b)

12. 96 – 4x – x²

Solution:

We have,

96 – 4x – x²

-x² – 4x + 96

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

-x² – 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)² + (a²)² + 2 (2a²) + 4 – a²

(a² + 2)² + (-a²)

By using the formula (a² – b²) = (a+b) (a-b)

(a² + 2 + a) (a² + 2 – a)

(a² + a + 2) (a² – a + 2)

14. 4x⁴ + 1

Solution:

We have,

4x⁴ + 1

(2x²)² + 1 + 4x² – 4x²

(2x² + 1)² – 4x²

By using the formula (a² – b²) = (a+b) (a-b)

(2x² + 1 + 2x) (2x² + 1 – 2x)

(2x² + 2x + 1) (2x² – 2x + 1)

15. 4x⁴ + y⁴

Solution:

We have,

4x⁴ + y⁴

(2x²)² + (y²)² + 4x²y² – 4x²y²

(2x² + y²)² – 4x²y²

By using the formula (a² – b²) = (a+b) (a-b)

(2x² + y² + 2xy) (2x² + y² – 2xy)

16. (x + 2)⁴ – 6(x + 2) + 9

Solution:

We have,

(x + 2)⁴ – 6(x + 2) + 9

(x² + 2²)² – 6x – 12 + 9

(x² + 2² + 2(2)(x)) – 6x – 12 + 9

x² + 4 + 4x – 6x – 12 + 9

x² – 2x + 1

By using the formula (a – b)² = a² + b² – 2ab

(x – 1)²

17. 25 – p² – q² – 2pq

Solution:

We have,

25 – p² – q² – 2pq

25 – (p² + q² + 2pq)

(5)² – (p + q)²

By using the formula (a² – b²) = (a+b) (a-b)

(5 + p + q) (5 –p – q)

-(p + q + 5) (p + q – 5)

18. x² + 9y² – 6xy – 25a²

Solution:

We have,

x² + 9y² – 6xy – 25a²

(x – 3y)² – (5a)²

By using the formula (a² – b²) = (a+b) (a-b)

(x – 3y + 5a) (x – 3y – 5a)

19. 49 – a² + 8ab – 16b²

Solution:

We have,

49 – a² + 8ab – 16b²

49 – (a² – 8ab + 16b²)

49 – (a – 4b)²

By using the formula (a² – b²) = (a + b) (a – b)

(7 + a – 4b) (7 – a + 4b)

-(a – 4b + 7) (a – 4b – 7)

20. a² – 8ab + 16b² – 25c²

Solution:

We have,

a² – 8ab + 16b² – 25c²

(a – 4b)²– (5c)²

By using the formula (a² – b²) = (a+b) (a-b)

(a – 4b + 5c) (a – 4b – 5c)

21. x² – y² + 6y – 9

Solution:

We have,

x² – y² + 6y – 9

x² + 6y – (y² – 6y + 9)

x² – (y – 3)²

By using the formula (a² – b²) = (a+b) (a-b)

(x + y – 3) (x – y + 3)

22. 25x² – 10x + 1 – 36y²

Solution:

We have,

25x² – 10x + 1 – 36y²

(5x)² – 2 (5x) + 1 – (6y)²

(5x – 1)² – (6y)²

By using the formula (a² – b²) = (a+b) (a-b)

(5x – 6y – 1) (5x + 6y – 1)

23. a² – b² + 2bc – c²

Solution:

We have,

a² – b² + 2bc – c²

a² – (b² – 2bc + c²)

a² – (b – c)²

By using the formula (a² – b²) = (a+b) (a-b)

(a + b – c) (a – b + c)

24. a² + 2ab + b² – c²

Solution:

We have,

a² + 2ab + b² – c²

(a + b)² – c²

By using the formula (a² – b²) = (a+b) (a-b)

(a + b + c) (a + b – c)

25. 49 – x² – y² + 2xy

Solution:

We have,

49 – x² – y² + 2xy

49 – (x² + y² – 2xy)

7² – (x – y)²

By using the formula (a² – b²) = (a+b) (a-b)

(x – y + 7) (y – x + 7)

26. a² + 4b² – 4ab – 4c²

Solution:

We have,

a² + 4b² – 4ab – 4c²

a² – 2 (a) (2b) + (2b)² – (2c)²

(a – 2b)² – (2c)²

By using the formula (a² – b²) = (a+b) (a-b)

(a – 2b + 2c) (a – 2b – 2c)

27. x² – y² – 4xz + 4z²

Solution:

We have,

x² – y² – 4xz + 4z²

x² – 2 (x) (2z) + (2z)² – y²

As (a-b)² = a² + b² – 2ab

(x – 2z)² – y²

By using the formula (a² – b²) = (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. x² + 12x – 45

Solution:

We have,

x² + 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

x² + 12x – 45 = x² + 15x – 3x – 45

= x (x + 15) – 3 (x + 15)

= (x – 3) (x + 15)

2. 40 + 3x – x²

Solution:

We have,

40 + 3x – x²

-(x² – 3x – 40)

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

So we can replace -3x with 5x – 8x

-40 by 5 × -8

-(x² – 3x – 40) = x² + 5x – 8x – 40

= -x (x + 5) – 8 (x + 5)

= -(x – 8) (x + 5)

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

3. a² + 3a – 88

Solution:

We have,

a² + 3a – 88

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

So we can replace 3a with 11a – 8a

-40 by -11 × 8

a² + 3a – 88 = a² + 11a – 8a – 88

= a (a + 11) – 8 (a + 11)

= (a – 8) (a + 11)

4. a² – 14a – 51

Solution:

We have,

a² – 14a – 51

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

So we can replace -14a with 3a – 17a

-51 by -17 × 3

a² – 14a – 51 = a² + 3a – 17a – 51

= a (a + 3) – 17 (a + 3)

= (a – 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 with 5x + 9x

45 by 5 × 9

x² + 14x + 45 = x² + 5x + 9x + 45

= x (x + 5) – 9 (x + 5)

= (x + 9) (x + 5)

6. x² – 22x + 120

Solution:

We have,

x² – 22x + 120

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

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

120 by -12 × -10

x² – 22x + 120 = x² – 12x – 10x + 120

= x (x – 12) – 10 (x – 12)

= (x – 10) (x – 12)

7. x² – 11x – 42

Solution:

We have,

x² – 11x – 42

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

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

-42 by 3 × -14

x² – 11x – 42 = x² + 3x – 14x – 42

= x (x + 3) – 14 (x + 3)

= (x – 14) (x + 3)

8. a² + 2a – 3

Solution:

We have,

a² + 2a – 3

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

So we can replace 2a with 3a -a

-3 by 3 × -1

a² + 2a – 3 = a² + 3a – a – 3

= a (a + 3) – 1 (a + 3)

= (a – 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 with 8a + 6a

48 by 8 × 6

a² + 14a + 48 = a² + 8a + 6a + 48

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

= (a + 6) (a + 8)

10. x² – 4x – 21

Solution:

We have,

x² – 4x – 21

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

So we can replace -4x with 3x – 7x

-21 by 3 × -7

x² + 4x – 21 = x² + 3x – 7x – 21

= x (x + 3) – 7 (x + 3)

= (x – 7) (x + 3)

11. y² + 5y – 36

Solution:

We have,

y² + 5y – 36

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

So we can replace 5y with 9y – 4y

-36 by 9 × -4

y² + 5y – 36 = y² + 9y – 4y – 36

= y (y + 9) – 4 (y + 9)

= (y – 4) (y + 9)

12. (a² – 5a)² – 36

Solution:

We have,

(a² – 5a)² – 36

(a² – 5a)² – 6²

By using the formula (a² – b²) = (a+b) (a-b)

(a² – 5a)² – 6² = (a² – 5a + 6) (a² – 5a – 6)

So now we shall factorize the expression (a² – 5a + 6)

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

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

6 by 1 × -6

a² -5a – 6 = a² + a – 6a – 6

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

= (a – 6) (a + 1)

So now we shall factorize the expression (a² – 5a + 6)

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

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

6 by -2 × -3

a² -5a + 6 = a² – 2a – 3a + 6

= a (a – 2) -3 (a – 2)

= (a – 3) (a – 2)

∴ (a² – 5a)² – 36 = (a² – 5a + 6) (a² – 5a – 6)

= (a + 1) (a – 6) (a – 2) (a – 3)

13. (a + 7) (a – 10) + 16

Solution:

We have,

(a + 7) (a – 10) + 16

a² – 10a + 7a – 70 + 16

a² – 3a – 54

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

So we can replace -3a with 6a – 9a

-54 by 6 × -9

a² – 3a – 54 = a² + 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. 2x² + 5x + 3

Solution:

We have,

2x² + 5x + 3

The coefficient of x² 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

2x² + 5x + 3 = 2x² + 2x + 3x + 3

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

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

2. 2x² – 3x – 2

Solution:

We have,

2x² – 3x – 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

2x² – 3x – 2 = 2x² – 4x + x – 2

= 2x (x – 2) + 1 (x – 2)

= (x – 2) (2x + 1)

3. 3x² + 10x + 3

Solution:

We have,

3x² + 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

3x² + 10x + 3 = 3x² + 9x + x + 3

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

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

4. 7x – 6 – 2x²

Solution:

We have,

7x – 6 – 2x²

– 2x² + 7x – 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

2x² – 7x + 6 = 2x² – 4x – 3x + 6

= 2x (x – 2) – 3 (x – 2)

= (x – 2) (2x – 3)

5. 7x² – 19x – 6

Solution:

We have,

7x² – 19x – 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 – 21x

7x² – 19x – 6 = 7x² + 2x – 21x – 6

= x (7x + 2) – 3 (7x + 2)

= (7x + 2) (x – 3)

6. 28 – 31x – 5x²

Solution:

We have,

28 – 31x – 5x²

– 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

5x² + 31x – 28 = 5x² – 4x + 35x – 28

= x (5x – 4) + 7 (5x – 4)

= (x + 7) (5x – 4)

7. 3 + 23y – 8y²

Solution:

We have,

3 + 23y – 8y²

– 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

8y² – 23y – 3 = 8y² – 24y + y – 3

= 8y (y – 3) + 1 (y – 3)

= (8y + 1) (y – 3)

8. 11x² – 54x + 63

Solution:

We have,

11x² – 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 – 21x

11x² – 54x + 63 = 11x² – 33x – 21x + 63

= 11x (x – 3) – 21 (x – 3)

= (11x – 21) (x – 3)

9. 7x – 6x² + 20

Solution:

We have,

7x – 6x² + 20

– 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

6x² – 7x – 20 = 6x² – 15x + 8x – 20

= 3x (2x – 5) + 4 (2x – 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

3x² + 22x + 35 = 3x² + 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

12x² -17xy+ 6y² = 12x² – 9xy – 8xy + 6y²

= 3x (4x – 3y) – 2y (4x – 3y)

= (3x – 2y) (4x – 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

6x² -5xy- 6y² = 6x² + 4xy – 9xy – 6y²

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

= (2x – 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 – xy

6x² -13xy+ 2y² = 6x² – 12xy – xy + 2y²

= 6x (x – 2y) – y (x – 2y)

= (6x – 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 – 10xy

14x² + 11xy- 15y² = 14x² + 21xy – 10xy – 15y²

= 2x (7x – 5y) + 3y (7x – 5y)

= (2x + 3y) (7x – 5y)

15. 6a² + 17ab – 3b²

Solution:

We have,

6a² + 17ab – 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 – ab

6a² +17ab– 3b² = 6a² + 18ab – ab – 3b²

= 6a (a + 3b) – b (a + 3b)

= (6a – b) (a + 3b)

16. 36a² + 12abc – 15b²c²

Solution:

We have,

36a² + 12abc – 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 – 18abc

36a² –12abc– 15b²c² = 36a² + 30abc – 18abc – 15b²c²

= 6a (6a + 5bc) – 3bc (6a + 5bc)

= (6a + 5bc) (6a – 3bc)

= (6a + 5bc) 3(2a – bc)

17. 15x² – 16xyz – 15y²z²

Solution:

We have,

15x² – 16xyz – 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

15x² -16xyz- 15y²z² = 15x² – 25yz + 9yz – 15y²z²

= 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. p² + 6p + 8

Solution:

We have,

p² + 6p + 8

The coefficient of p² is unity. So, we add and subtract the square of half of the coefficient of p.

p² + 6p + 8 = p² + 6p + 3² – 3² + 8 (Adding and subtracting 3²)

= (p + 3)² – 1² (By completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (p + 3 – 1) (p + 3 + 1)

= (p + 2) (p + 4)

2. q² – 10q + 21

Solution:

We have,

q² – 10q + 21

The coefficient of q² is unity. So, we add and subtract the square of half of the coefficient of q.

q² – 10q + 21 = q² – 10q+ 5² – 5² + 21 (Adding and subtracting 5²)

= (q – 5)² – 2² (By completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (q – 5 – 2) (q – 5 + 2)

= (q – 3) (q – 7)

3. 4y² + 12y + 5

Solution:

We have,

4y² + 12y + 5

4(y² + 3y + 5/4)

The coefficient of y² is unity. So, we add and subtract the square of half of the coefficient of y.

4(y² + 3y + 5/4) = 4 [y² + 3y + (3/2)² – (3/2)² + 5/4] (Adding and subtracting (3/2)²)

= 4 [(y + 3/2)² – 1²] (Completing the square)

By using the formula (a² – b²) = (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. p² + 6p – 16

Solution:

We have,

p² + 6p – 16

The coefficient of p² is unity. So, we add and subtract the square of half of the coefficient of p.

p² + 6p – 16 = p² + 6p + 3² – 3² – 16 (Adding and subtracting 3²)

= (p + 3)² – 5² (Completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (p + 3 + 5) (p + 3 – 5)

= (p + 8) (p – 2)

5. x² + 12x + 20

Solution:

We have,

x² + 12x + 20

The coefficient of x² is unity. So, we add and subtract the square of half of the coefficient of x.

x² + 12x + 20 = x² + 12x + 6² – 6² + 20 (Adding and subtracting 6²)

= (x + 6)² – 4² (Completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (x + 6 + 4) (x + 6 – 4)

= (x + 2) (x + 10)

6. a² – 14a – 51

Solution:

We have,

a² – 14a – 51

The coefficient of a² is unity. So, we add and subtract the square of half of the coefficient of a.

a² – 14a – 51 = a² – 14a + 7² – 7² – 51 (Adding and subtracting 7²)

= (a – 7)² – 10² (Completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (a – 7 + 10) (9 – 7 – 10)

= (a – 17) (a + 3)

7. a² + 2a – 3

Solution:

We have,

a² + 2a – 3

The coefficient of a² is unity. So, we add and subtract the square of half of the coefficient of a.

a² + 2a – 3 = a² + 2a + 1² – 1² – 3 (Adding and subtracting 1²)

= (a + 1)² – 2² (Completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (a + 1 + 2) (a + 1 – 2)

= (a + 3) (a – 1)

8. 4x² – 12x + 5

Solution:

We have,

4x² – 12x + 5

4(x² – 3x + 5/4)

The coefficient of x² is unity. So, we add and subtract the square of half of the coefficient of x.

4(x² – 3x + 5/4) = 4 [x² – 3x + (3/2)² – (3/2)² + 5/4] (Adding and subtracting (3/2)²)

= 4 [(x – 3/2)² – 1²] (Completing the square)

By using the formula (a² – b²) = (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. y² – 7y + 12

Solution:

We have,

y² – 7y + 12

The coefficient of y² is unity. So, we add and subtract the square of half of the coefficient of y.

y² – 7y + 12 = y² – 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 (a² – b²) = (a+b) (a-b)

= (y – (7/2- 1/2)) (y – (7/2 + 1/2))

= (y – 3) (y – 4)

10. z² – 4z – 12

Solution:

We have,

z² – 4z – 12

The coefficient of z² is unity. So, we add and subtract the square of half of the coefficient of z.

z² – 4z – 12 = z² – 4z + 2² – 2² – 12 [Adding and subtracting 2²]

= (z – 2)² – 4² (Completing the square)

By using the formula (a² – b²) = (a+b) (a-b)

= (z – 2 + 4) (z – 2 – 4)

= (z – 6) (z + 2)