\(p(x)={ x }^{ 3 }-{ 4x }^{ 2 }+x+6, g(x)=x-3\)
g(x) = 0
\(\Rightarrow x-3=0\ \Rightarrow \ x=3\)
\(\therefore\) Zero of g(x) is 3
Now, p(3),
\({ =(3) }^{ 3 }-{ 4(3) }^{ 2 }+3+6\)
\(\\ =27-36+3+6=0\)
\(\therefore\) By factor theorem, g(x) is a factor of p(x)

\(x-\frac{1}{2}\)
Remainder
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 3 }+{ 3\left( \frac { 1 }{ 2 } \right) }^{ 2 }+3{ \left( \frac { 1 }{ 2 } \right) }+1\\ =\frac { 1 }{ 8 } +\frac { 3 }{ 4 } +\frac { 3 }{ 2 } +1=\frac { 27 }{ 8 } \)

\(\therefore \ p(0)=2+0+2{ (0) }^{ 2 }-{ (0) }^{ 3 }=2\)
\(\\ p(1)=2+1+2{ (1) }^{ 2 }-{ (1) }^{ 3 }\)
\(\\ =2+1+2-1=4\)
and \(p(2)=2+2+2{ (2) }^{ 2 }-{ (2) }^{ 3 }\)
\(\\ =2+2+8-8=4\)

This expression is not a polynomial because in the term \(\frac { 2 }{ y } \) , the exponent of y is (- 1) which is not a whole number.

(x+2)(x-3)(2x+1)

As x – 1 is a factor of p(x) = 4x³ + 3x² – 4x + k, p(1) = 0
Now, p(1) = 4(1)³ + 3(1)² – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3

2

\(\frac { 1 }{ 4 } \)

(i) We know that if x + y + z = 0, then 
\({ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }=3xyz\)
\(\therefore 27{ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\)
 \(={ (3x) }^{ 2 }+{ (y) }^{ 3 }+{ (z) }^{ 3 }\)
= 3(3x)(y)(z)  | \(\therefore\) 3x + y + z = 0
= 9xyz
(ii) Clearly each of x, y, and z is a factor of \(27{ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\)
(ii) We have seen that 9 is the factor of \(27{ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\) and 9 is a perfect square. hence the result. So, Meenu is correct. So, the value 'Expertness' is depicted by her finding.
(iv) The mathematical concept ' Polynomials' has been covered in this problem.
(v) The formulae used in the solution are as follow:
1. If x + y + z = 0, then, \(27{ x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 }\)= 3xyz
2. Concept of factors
3. Concept of perfect square numbers.

Definition of zero polynomial

Convention

(i) Here p(x) = x² + x + k
For x - 1 be a factor of p(x), p(1) should be equal to 0.
We have p(1) = (1)² + 1 + k
or p(1) = 1 + 1 + k = k + 2
∴ k+2 = 0
⇒ k = -2
Here, p(x) = 2x² + kx + \(\sqrt{2}\)
For x - 1 be a factor of p(x), p(1) = 0
Since, p(1) = 2(1)² + k(1) +  \(\sqrt{2}\)
=2 + k+  \(\sqrt{2}\)
∵ p(1) must be equal to 0.
∴ k + 2 +  \(\sqrt{2}\)=0
⇒ k = -2 -  \(\sqrt{2}\)
or k = - (2 +  \(\sqrt{2}\)) .
(iii) Here p(x) = kx² -  \(\sqrt{2}\)x + 1 and g(x) = x - 1
∴ For (x - I) be a factor of p(x), p(1) should be equal to 0.
Since p(1) = k(1)² -  \(\sqrt{2}\)(1) + 1
or p( 1) = k -  \(\sqrt{2}\) + 1
or p( 1) = k -  \(\sqrt{2}\) + 1
∴ k -  \(\sqrt{2}\) + 1 = 0
⇒ k =  \(\sqrt{2}\) - 1
(iv) Here p(x) = kx² - 3x + k and g(x) = x - 1
For g(x) be a factor of p(x), p(1) should be equal to 0.
Since p(1) = k(1)² - 3(1) + k
= k - 3 + k
= 2k - 3
2k - 3 = 0
k = \(\frac{3}{2}\)

Here, (i) and (ii) are the polynomials in one variable, (v) is a polynomial in three variables and (iii) and (iv) are not polynomials, because in each of these exponents of the variable is not a whole number.