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# Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.9

• Difficulty Level : Basic
• Last Updated : 01 Dec, 2020

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

### Question 1. p2 + 6p + 8

Solution:

p2 + 6p + 8

= p2 + 2 x p x 3 + 32 – 32 + 8   (completing the square)

= (p2 + 6p + 32) – 1

= (p + 3)2 – 1

= (p + 3)2 – (1)2       { ∵ a2 + b2  = (a + b) (a – b)}

= (p + 3 + 1) (p + 3 – 1)

= (p + 4) (p + 2)

### Question 2. q2 – 10q + 21

Solution:

q2 – 10q + 21

= (q)2 – 2 x q x 5 + (5)2 – (5)2 + 21   (completing the square)

= (q)2 – 2 x q x 5 + (5)2 – 25+ 21

= (q)2 – 2 x q x 5 + (5)2 – 25 +21

= (q)2 – 2 x q x 5 + (5)2 – 4

= (q – 5)2 – (2)     {∵ a2 – b2 = (a + b) (a – b)}

= (q – 5 + 2) (q – 5 – 2)

=(q – 3) (q – 7)

### Question 3. 4y2 + 12y + 5

Solution:

4y2 + 12y + 5

= (2y)2 + 2 x 2y x 3 + (3)2 – (3)2 + 5    (completing the square)

= (2y + 3)2 – 9 + 5

= (2y + 3)2 – 4

= (2y + 3)2 – (2)2   {∵ a2 – b2 = (a + b) (a – b)}

= (2y + 3 + 2) (2y + 3 – 2)

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

### Question 4. p2 + 6p – 16

Solution:

p2 + 6p – 16

= (p)2 + 2 x  p x 3 + (3)2 – (3)2 – 16    (completing the square)

= (p)2 + 2 x p x 3 + (3)2 – 9 – 16

= (p + 3)2 – 25

= (p + 3)2 – (5)2     {∵ a2 – b2 = {a + b) (a – b)}

= (p + 3 + 5)(p + 3 – 5)

= (p + 8) (p – 2)

### Question 5. x2 + 12x + 20

Solution:

x2 + 12x + 20

= (x)2 + 2 x x x 6 + (6)2 – (6)2 + 20   (completing the square)

= (x)2 + 2 x x x 6 + (6)2 -36 + 20

= (x + 6)2 -16

= (x + 6)2 – (4)2   {∵ a2 – b2 = (a + b) (a – b)}

= (x + 6 + 4) (x + 6 – 4)

= (x + 10) (x + 2)

### Question 6. a2 – 14a – 51

Solution:

a2 – 14a – 51

= (a)2 – 2 x a x 7 + (7)2 – (7)2 – 51       (completing the square)

= (a)2 – 2 x a x 7 + (7)2 – 49 – 51

= (a – 7)2 – 100

= (a – 7)2 – (10)2    {∵  a2 – b2 = (a + b) (a – b)}

= (a – 7 + 10) (a – 7 – 10)

= (a + 3) (a – 17)

### Question 7. a2 + 2a – 3

Solution:

a2 + 2a – 3

= (a)2 + 2 x a x 1 + (1)2 – (1)2 – 3   (completing the square)

= (a)2 + 2 x a x 1 + (1)2 – 1 – 3

= (a + 1)2 – 4

= (a + 1)2 – (2)2 {∵ a2 – b2 = (a + b) (a – b)}

= (a + 1 + 2) (a + 1 – 2)

= (a + 3) (a – 1)

### Question 8. 4x2 – 12x + 5

Solution:

4x2 – 12x + 5

= (2x)2 – 2 x 2x x 3 + (3)2 – (3)2 + 5  (completing the square)

= (2x)2 – 2 x 2x x 3 + (3)2 – 9 + 5

= (2x – 3)2 – 4

= (2x – 3)2 – (2)2      {∵ a2 – b2 = (a + b) (a – b)}

= (2x – 3 + 2) (2x – 3 – 2)

= (2x – 1) (2x – 5)

### Question 9. y2 – 7y + 12

Solution:

y2 – 7y + 12

= (y)2 – 2 × y ×  7/2 + (7/2)2 – (7/2)2 + 12      (completing the square)

= (y)2 – 2 × y × 7/2 + 49/4 – 49/4 + 12

= (y – 7/2)2 – (49 – 48)/4

= (y – 7/2)2 – 1/4

= (y – 7/2)2 – (1/2)2         {∵ a2 – b2 = (a + b) (a – b)}

= (y – 7/2 + 1/2) (y – 7/2 – 1/2)

= (y – 6/2)  (y – 8/2)

= (y – 3) (y – 4)

### Question 10. z2 – 4z -12

Solution:

z2 – 4z – 12

= (z)2 – 2 x z x 2 + (2)2 – (2)2 – 12  (completing the square)

= (z)2 – 2 x z x 2 + (2)2 – 4 – 12

= (z – 2)2 – 16

= (z – 2)2– (4)2   {∵ a2 – b2 = (a + b) (a – b)}

= (z – 2 + 4) (z – 2 – 4)

= (z + 2)(z – 6)

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