#### (i) (x + 2y + 3) (x + 2y + 7)

#### (ii) (2x + y + 5) (2x + y – 9)

#### (iii) (x – 2y – 5) (x – 2y + 3)

#### (iv) (3x – 4y – 2) (3x – 4y – 6)

**Answer :**

**(i) (x + 2y + 3) (x + 2y + 7)**

Consider x + 2y = a

(a + 3) (a + 7) = a^{2} + (3 + 7) a + 3 × 7

By further calculation

= a^{2} + 10a + 21

Substituting the value of a

= (x + 2y)^{2} + 10 (x + 2y) + 21

By expanding using formula

= x^{2} + 4y^{2} + 2 × x × 2y + 10x + 20y + 21

So we get

= x^{2} + 4y^{2} + 4xy + 10x + 20y + 21

**(ii) (2x + y + 5) (2x + y – 9)**

Consider 2x + y = a

(a + 5) (a – 9) = a^{2} + (5 – 9) a + 5 × (-9)

By further calculation

= a^{2} – 4a – 45

Substituting the value of a

= (2x + y)^{2} – 4 (2x + y) – 45

By expanding using formula

= 4x^{2} + y^{2} + 2 × 2x × y – 8x – 4y – 45

So we get

= 4x^{2} + y^{2} + 4xy – 8x – 4y – 45

**(iii) (x – 2y – 5) (x – 2y + 3)**

Consider x – 2y = a

(a – 5) (a + 3) = a^{2} + (- 5 + 3) a + (-5) (3)

By further calculation

= a^{2} – 2a – 15

Substituting the value of a

= (x – 2y)^{2} – 2 (x – 2y) – 15

By expanding using formula

= x^{2} + 4y^{2} – 2 × x × 2y – 2x + 4y – 15

So we get

= x^{2} + 4y^{2} – 4xy – 2x + 4y – 15

**(iv) (3x – 4y – 2) (3x – 4y – 6)**

Consider 3x – 4y = a

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

By further calculation

= a^{2} – 8a + 12

Substituting the value of a

= (3x – 4y)^{2} – 8 (3x – 4y) + 12

Expanding using formula

= 9x^{2} + 16y^{2} – 2 × 3x × 4y – 24x + 32y + 12

So we get

= 9x^{2} + 16y^{2} – 24xy – 24x + 32y + 12

**More Solutions:**

- Trigonometric Ratios of Standard Angles
- Find the value of A if
- Find the value of θ (0° < θ < 90°) if:
- If A, B and C are the interior angles of a △ ABC
- The value of tan 30/cot60 is
- The value of (sin 45 + cos 45 ) is
- The value of tan² 30 – 4 sin² 45 is
- If A = 30, then the value of 2 sin A Cos A is
- The value of (sin 30 + cos 30) – (sin 60 + cos 60) is
- The value of √3 cosec 60 – sec 60 is