When y = 0
f(y) = 6y-3y² + 3
f(0) = 6(0) - 3(0)²+ 3
= 0 - 0+ 3 = 3

When y = -1
f(y) = 6y - 3y² + 3
f(-1) = 6(-1) - 3(-1)²+ 3
= -6 - 3+ 3 = -6

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)

Given that f(x) = 2x + 1 = \(2\left( x+\frac { 1 }{ 2 } \right) \)
\(=2\left( x-\left( -\frac { 1 }{ 2 } \right) \right) \)
\(f\left( -\frac { 1 }{ 2 } \right) =2\left[ -\frac { 1 }{ 2 } -\left( -\frac { 1 }{ 2 } \right) \right] =2\left( 0 \right) =0\)
Since \(f\left( -\frac { 1 }{ 2 } \right) =0,\ x=-\frac { 1 }{ 2 } \)is the zero of f(x)

f(x) = x² - 4x + 3

Since the value of the polynomial f(x) at x = 1 and x = 3 is zero, as the zeros of polynomial f(x) are 1 and 3.

Coefficient of x² is -2 and Coefficient of x is  -\(\sqrt 7\)

\({ m }^{ 2 }-\sqrt [ 3 ]{ m } +7m-10\) not a polynomial.
Since the exponent of m is not a whole number \(\left( \sqrt [ 3 ]{ m } ={ m }^{ \frac { 1 }{ 3 } } \right) \) 

\(\frac { 1 }{ { x }^{ -2 } } +\frac { 1 }{ { x }^{ -1 } } +7\) is a polynomial (Since \(\frac { 1 }{ { x }^{ -2 } } ={ x }^{ 2 }\) and \(\frac { 1 }{ { x }^{ -1 }=x } \) = x)

x²(x -1) is a polynomial
non negative integral power.