Answer :
Certainly! Let's go step by step to factorise each given polynomial and solve the additional problem about the area of the rectangular field. We are provided with several polynomial expressions to factorise directly.
### Part 1: Factorising the Polynomials
a) Given polynomial: [tex]\( x^2 + 4x + 3 \)[/tex]
[tex]\[ \boxed{(x + 1)(x + 3)} \][/tex]
b) Given polynomial: [tex]\( x^2 - 7x - 8 \)[/tex]
[tex]\[ \boxed{(x - 8)(x + 1)} \][/tex]
c) Given polynomial: [tex]\( a^2 - 27a + 180 \)[/tex]
[tex]\[ \boxed{(a - 15)(a - 12)} \][/tex]
d) Given polynomial: [tex]\( 2x^2 + 7x + 6 \)[/tex]
[tex]\[ \boxed{(x + 2)(2x + 3)} \][/tex]
e) Given polynomial: [tex]\( 3p^2 - 7p - 6 \)[/tex]
[tex]\[ \boxed{(p - 3)(3p + 2)} \][/tex]
f) Given polynomial: [tex]\( 2x^2 + 3xy - 5y^2 \)[/tex]
[tex]\[ \boxed{(x - y)(2x + 5y)} \][/tex]
g) Given polynomial: [tex]\( 3a^2 - 16ab + 13b^2 \)[/tex]
[tex]\[ \boxed{(a - b)(3a - 13b)} \][/tex]
h) Given polynomial: [tex]\( 9a^3bx + 12a^2b^2x - 5ab^3x \)[/tex]
[tex]\[ \boxed{a \cdot b \cdot x \cdot (3a - b)(3a + 5b)} \][/tex]
i) Given polynomial: [tex]\( 12\frac{a^2}{b^2} + \frac{a}{b} - 20 \)[/tex]
[tex]\[ \boxed{\frac{(3a + 4b)(4a - 5b)}{b^2}} \][/tex]
j) Given polynomial: [tex]\( \frac{x^2}{y^2} - 2 - \frac{3y^2}{x^2} \)[/tex]
[tex]\[ \boxed{\frac{(x^2 - 3y^2)(x^2 + y^2)}{x^2 y^2}} \][/tex]
k) Given polynomial: [tex]\( 2(x + y)^2 + 9(x + y) + 7 \)[/tex]
[tex]\[ \boxed{(x + y + 1)(2x + 2y + 7)} \][/tex]
l) Given polynomial: [tex]\( 3(x - y)^2 - 10(x - y) + 8 \)[/tex]
[tex]\[ \boxed{(x - y - 2)(3x - 3y - 4)} \][/tex]
### Part 2: Rectangular Field Problem
The area of a rectangular field is given by [tex]\( x^2 + 8x + 15 \)[/tex] sq. m.
(i) Find the length and breadth of the field:
[tex]\[ x^2 + 8x + 15 = (x + 3)(x + 5) \][/tex]
Therefore, the length and breadth of the field are [tex]\( (x + 3) \)[/tex] meters and [tex]\( (x + 5) \)[/tex] meters, respectively.
[tex]\[ \boxed{(x + 3) \text{ and } (x + 5) \text{ meters }} \][/tex]
(ii) Find the perimeter of the field:
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2(\text{length} + \text{breadth}) \][/tex]
Substituting the length and breadth:
[tex]\[ P = 2((x + 3) + (x + 5)) = 2(2x + 8) = 4x + 16 \text{ meters } \][/tex]
[tex]\[ \boxed{4x + 16 \text{ meters}} \][/tex]
### Part 1: Factorising the Polynomials
a) Given polynomial: [tex]\( x^2 + 4x + 3 \)[/tex]
[tex]\[ \boxed{(x + 1)(x + 3)} \][/tex]
b) Given polynomial: [tex]\( x^2 - 7x - 8 \)[/tex]
[tex]\[ \boxed{(x - 8)(x + 1)} \][/tex]
c) Given polynomial: [tex]\( a^2 - 27a + 180 \)[/tex]
[tex]\[ \boxed{(a - 15)(a - 12)} \][/tex]
d) Given polynomial: [tex]\( 2x^2 + 7x + 6 \)[/tex]
[tex]\[ \boxed{(x + 2)(2x + 3)} \][/tex]
e) Given polynomial: [tex]\( 3p^2 - 7p - 6 \)[/tex]
[tex]\[ \boxed{(p - 3)(3p + 2)} \][/tex]
f) Given polynomial: [tex]\( 2x^2 + 3xy - 5y^2 \)[/tex]
[tex]\[ \boxed{(x - y)(2x + 5y)} \][/tex]
g) Given polynomial: [tex]\( 3a^2 - 16ab + 13b^2 \)[/tex]
[tex]\[ \boxed{(a - b)(3a - 13b)} \][/tex]
h) Given polynomial: [tex]\( 9a^3bx + 12a^2b^2x - 5ab^3x \)[/tex]
[tex]\[ \boxed{a \cdot b \cdot x \cdot (3a - b)(3a + 5b)} \][/tex]
i) Given polynomial: [tex]\( 12\frac{a^2}{b^2} + \frac{a}{b} - 20 \)[/tex]
[tex]\[ \boxed{\frac{(3a + 4b)(4a - 5b)}{b^2}} \][/tex]
j) Given polynomial: [tex]\( \frac{x^2}{y^2} - 2 - \frac{3y^2}{x^2} \)[/tex]
[tex]\[ \boxed{\frac{(x^2 - 3y^2)(x^2 + y^2)}{x^2 y^2}} \][/tex]
k) Given polynomial: [tex]\( 2(x + y)^2 + 9(x + y) + 7 \)[/tex]
[tex]\[ \boxed{(x + y + 1)(2x + 2y + 7)} \][/tex]
l) Given polynomial: [tex]\( 3(x - y)^2 - 10(x - y) + 8 \)[/tex]
[tex]\[ \boxed{(x - y - 2)(3x - 3y - 4)} \][/tex]
### Part 2: Rectangular Field Problem
The area of a rectangular field is given by [tex]\( x^2 + 8x + 15 \)[/tex] sq. m.
(i) Find the length and breadth of the field:
[tex]\[ x^2 + 8x + 15 = (x + 3)(x + 5) \][/tex]
Therefore, the length and breadth of the field are [tex]\( (x + 3) \)[/tex] meters and [tex]\( (x + 5) \)[/tex] meters, respectively.
[tex]\[ \boxed{(x + 3) \text{ and } (x + 5) \text{ meters }} \][/tex]
(ii) Find the perimeter of the field:
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2(\text{length} + \text{breadth}) \][/tex]
Substituting the length and breadth:
[tex]\[ P = 2((x + 3) + (x + 5)) = 2(2x + 8) = 4x + 16 \text{ meters } \][/tex]
[tex]\[ \boxed{4x + 16 \text{ meters}} \][/tex]
