p(x₁) = 2x³ + ax² + 4x - 12
When it is divided by x - 3,
p(3) = 2(3)³ + a(3)² + 4(3) - 12
= 54 + 9a + 12 - 12
= 54 + 9a .....(R₁)
p(x2) = x³ + x² - 2x + a
When it is divided by x - 3,
p(3) = 3³ + 3² - 2(3) + a
= 27 + 9 -6 +a
= 30 + a  ......(R₂)
The given remainders are same (R₁ = R₂)
\(\therefore\) 54 + 9a = 30 + a
9a - a = 30 - 54
8a = -24
\(\therefore\)  a = - 24/8 = - 3
Consider R₂,
Remainder = 30 - 3
= 27

3x³ - 2x² + 6x + 15 - (2x³ + 6x² - 5x + 8)
= 3x³ - 2x² + 6x + 15 - 2x³ - 6x² + 5x - 8
= 3x³ - 2x³- 2x² - 6x² + 6x + 5x + 15 - 8
= x³ - 8x² + 11x + 7
x³ - 8x² + 11x + 7 must be added to get 3x³ - 2x² + 6x + 15

The standard form is \(-\frac { 7 }{ 2 } { x }^{ 4 }-5{ x }^{ 3 }+\sqrt { 2 } { x }^{ 2 }+x\) or \(x+\sqrt { 2 } { x }^{ 2 }-{ 5x }^{ 3 }-\frac { 7 }{ 2 } { x }^{ 4 }\)

We see that p(x) + q(x) is also a polynomial. Hence the sum of any two polynomials is also a polynomial

p(x) = x² - \(2\sqrt { 2 } +1\)
\(p\left( 2\sqrt { 2 } \right) ={ \left( 2\sqrt { 2 } \right) }^{ 2 }-2\sqrt { 2 } \left( 2\sqrt { 2 } \right) +1\)
= 8 - 8 + 1
= 0 + 1 = 1

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)

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