64m³ + 27n³ = (4m)³ + (3n)³
= (4m + 3n) ((4m)² - (4m) (3n) + (3n)²)               [\(\because \) a³ + b³ = (a+b) (a² - ab + b²)]
= (4m + 3n)(16m² - 12mn + 9n²)

x⁴ - 9x² = x² (x² - 9) = x² (x² - 32) = x² (x - 3)(x + 3)

25m^-2 - 16n² = (5m)² - (4n)²             [ \(\because \) a²-b² = (a - b)(a + b)]
= (5m - 4n) (5m+ 4n)

(3x + 2) units

Quotient: 5x⁴ + 2x³ + 6x² + 3x +9
Remainder: 7

2x² - 4xy + 9y² + 5x - 9y + 3

p(x) = x³- 3x²+ 4x+ 50
g(x) = x - 3 
p(3) = 33 - 3(3)² + 4(3) + 50
=  27 - 27 + 12 + 50 = 62
\(\therefore\) The Remainder 62

Number of zeros = 2 (The curve is intersecting the x-axis at 2 points)

We have
\(\frac{x^{3}-4x^{2}+6x}{x}=\frac{x^{3}}{x}-\frac{4x^{2}}{x}+\frac{6x}{x} , x\ne0\)
= x²- 4x + 6

p(x) = 2x - 1;  x = \(\frac { 1 }{ 2 } \)
\(p\left( \frac { 1 }{ 2 } \right) =2\left( \frac { 1 }{ 2 } \right) -1\)
= 1 - 1
= 0
\(\therefore\) \(\frac { 1 }{ 2 } \) is the zero of the polynomial

9x - 4 = 0
9x = 4
x = \(\frac { 4 }{ 9 } \)
\(\frac { 4 }{ 9 } \) is the root of the polynomial

q(y) = 2y - 3
= \(2\left( y-\frac { 2 }{ 3 } \right) \)
\(q\left( \frac { 3 }{ 2 } \right) =2\left( \frac { 3 }{ 2 } -\frac { 3 }{ 2 } \right) \)
= 2 \(\times\) 0 = 0
Hence \(\frac { 3 }{ 2 } \) is the zero of f(z)

P(x) = 2x +5
= \(2\left( x+\frac { 5 }{ 2 } \right) \)
\(p\left( -\frac { 5 }{ 2 } \right) =2\left( -\frac { 5 }{ 2 } \right) +\frac { 5 }{ 2 } \)
= -5 + 5 = 2(0) = 0
Hence \(-\frac { 5 }{ 2 } \) is the zero of p(x)

5x – 3 = 0 (or) 5x = 3
Then, x = \(\frac{3}{5}\)

Given that f(x) = 3x – 5 = \(\left( x-\frac { 5 }{ 3 } \right) \)
\(f\left( \frac { 5 }{ 3 } \right) =3\left( \frac { 5 }{ 3 } -\frac { 5 }{ 3 } \right) =3\left( 0 \right) =0\)
Since \(f\left(\frac{5}{3}\right)=0\), \(x=\frac { 5 }{ 3 } \) is the zero of f(x)