(i) Define an Auxiliary Polynomial

Observe that for , the function values follow the pattern . We can define a new polynomial such that:

Since is a polynomial of degree 5 with a leading coefficient of is also a degree 5 polynomial with a leading coefficient of 1 . From the given conditions, . Thus, the roots of are , and 5 . We can write as:

Therefore, the original polynomial is:

Solve for (Question 12)

To find , substitute into the equation:

(ii) The sum of the roots of a polynomial is given by .

Expanding :

The coefficient of is the negative sum of the roots: .

Since , the term does not affect the coefficient of in a degree 5 polynomial.

Thus,

Sum of roots .

(iii) The product of the roots for a degree 5 polynomial is given by (where is the constant term).

First, find the constant term of :

The constant term of is .

Substituting this into :

The constant term is -114 .

Product of roots .

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