Since, any odd positive integer n is of the form 4m + 1 or 4m + 3.

if n = 4m + 1
n² = (4m + 1)²
= 16m² + 8m + 1
= 8(2m² + m) + 1
So n² = 8q + 1 ........... (i)) (where q = 2m² + m is a positive integer)

If n = (4m + 3)
n² = (4m + 3)²
= 16m² + 24m + 9
= 8(2m² + 3m + 1) + 1
So n² = 8q + 1 ...... (ii) (where q = 2m² + 3m + 1 is a positive integer)

From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.

if -2 and -1 are zeros of f(x) = 2x⁴ + x³ - 14x² - 19x - 6
x+2 and x+1 are factors of f(x)
So (x + 2)(x + 1) = x² + x + 2x + 2 = x² + 3x + 2 is a factor of f(x)

On long division of f(x) by x² + 3x + 2 we get

f(x) = 2x⁴ + x³ - 14x² - 19x - 6 = (2x² - 5x - 3)(x² + 3x + 2)
= (2x + 1)(x - 3)(x + 2)(x + 1)
Therefore, zeroes of the polynomial = {tex}\frac{{ - 1}}{2}{/tex}, 3, -2, -1.

Since the point (3, a) lies on the line 2x - 3y = 5, we have
2 {tex}\times{/tex} 3 - 3 {tex}\times{/tex} a = 5
{tex}\Rightarrow{/tex} 6 - 3a = 5
{tex}\Rightarrow{/tex} 3a = 1
{tex}\Rightarrow{/tex} {tex}a = \frac { 1 } { 3 }{/tex}

Since, any odd positive integer n is of the form 4m + 1 or 4m + 3.

if n = 4m + 1
n² = (4m + 1)²
= 16m² + 8m + 1
= 8(2m² + m) + 1
So n² = 8q + 1 ........... (i)) (where q = 2m² + m is a positive integer)

If n = (4m + 3)
n² = (4m + 3)²
= 16m² + 24m + 9
= 8(2m² + 3m + 1) + 1
So n² = 8q + 1 ...... (ii) (where q = 2m² + 3m + 1 is a positive integer)

From (i) and (ii) we conclude that the square of an odd positive integer is of the form 8q + 1, for some integer q.

Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408 {tex}\times{/tex} 2 + 216.
Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as
408 = 216 {tex}\times{/tex} 1 + 192.
Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192
216 = 192 {tex}\times{/tex} 1 + 24.
Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24 
192 = 24 × 8 + 0.
Now the  remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.
Therefore,
24 = 1032m - 408 {tex}\times{/tex} 5
1032m = 24 + 408 {tex}\times{/tex} 5
1032m = 24 + 2040
1032m = 2064  
{tex}m = \frac{{2064}}{{1032}}{/tex}
Therefore, m = 2.

The prime factors of:
378 = 2 {tex} \times{/tex} 3³ {tex}\times{/tex} 7
180 = 2³ {tex}\times{/tex} 3² {tex}\times{/tex} 5
420 = 2² {tex}\times{/tex} 3 {tex}\times{/tex} 7{tex}\times{/tex} 5
HCF = 2 {tex}\times{/tex} 3=6
and LCM(378, 180, 420) = 2² {tex}\times{/tex} 3³ {tex}\times{/tex} 5 {tex}\times{/tex} 7 = 3780
HCF {tex}\times{/tex} LCM = 6 {tex}\times{/tex} 3780 = 22680
Product of given numbers is = 378 {tex}\times{/tex} 180 {tex}\times{/tex} 420 = 28576800
Hence, HCF {tex}\times{/tex} LCM {tex}\ne{/tex} Product of three numbers.

if -2 and -1 are zeros of f(x) = 2x⁴ + x³ - 14x² - 19x - 6
x+2 and x+1 are factors of f(x)
So (x + 2)(x + 1) = x² + x + 2x + 2 = x² + 3x + 2 is a factor of f(x)

On long division of f(x) by x² + 3x + 2 we get

f(x) = 2x⁴ + x³ - 14x² - 19x - 6 = (2x² - 5x - 3)(x² + 3x + 2)
= (2x + 1)(x - 3)(x + 2)(x + 1)
Therefore, zeroes of the polynomial = {tex}\frac{{ - 1}}{2}{/tex}, 3, -2, -1.

Length of the  courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (therefore, 1 m = 100 cm)
Breadth of the courtyard = 13 in 20 cm = 1300 cm + 20 cm = 1320 cm
To find the square tile of maximum side we take the HCF of 1872 and 1320
Prime factorization of 1872 = 2⁴ {tex}\times{/tex} 3² {tex}\times{/tex} 13
Prime factorization of 1320 = 2³ {tex}\times{/tex} 3 {tex}\times{/tex} 5 {tex}\times{/tex} 11
HCF of 1872 and 1320 = 2³ {tex}\times{/tex} 3 = 24
Therefore, Length of side of the square tile = 24 cm
Number of tiles required = Area of the courtyard / Area of each tile = (Length {tex}\times{/tex} Breadth) /Side²
= (1872 cm {tex}\times{/tex} 1320 cm) ÷ (24 cm {tex}\times{/tex} 24 cm) = 4290.
Hence 4290 tiles are required.