Answer the following and justify:
(i) Can \(x^2-1\) be the quotient on division of \(x^6+2x^3+x-1\) by a polynomial in \(x\) of degree 5?
(ii) What will the quotient and remainder be on division of \(ax^2+bx+c\) by \(px^3+qx^2+rx+s\), \(p\ne0\)?
(iii) If on division of a polynomial \(p(x)\) by a polynomial \(g(x)\), the quotient is zero, what is the relation between \(\deg p\) and \(\deg g\)?
(iv) If on division of a non-zero polynomial \(p(x)\) by a polynomial \(g(x)\), the remainder is zero, what is the relation between \(\deg p\) and \(\deg g\)?
(v) Can the quadratic polynomial \(x^2+kx+k\) have equal zeroes for some odd integer \(k>1\)?
(i) No (ii) Quotient 0, remainder \(ax^2+bx+c\) (iii) \(\deg p<\deg g\) (iv) \(\deg p\ge \deg g\) (v) No
(i) If divisor has degree 5, then \(\deg\)(quotient) must be \(\deg\)dividend \(-\) \(\deg\)divisor \(=6-5=1\) (or less if leading terms cancel), not 2. Hence \(x^2-1\) cannot be the quotient.
(ii) Since \(\deg(px^3+\cdots)=3>\deg(ax^2+bx+c)=2\), the division yields quotient 0 and remainder the dividend itself.
(iii) Quotient 0 means divisor’s degree exceeds dividend’s: \(\deg p<\deg g\).
(iv) Remainder 0 means \(g(x)\) divides \(p(x)\). Thus \(\deg p\ge \deg g\) (allowing equality if leading coefficients match to cancel exactly).
(v) Equal zeroes require discriminant zero: \(k^2-4k=0\Rightarrow k(k-4)=0\Rightarrow k=0\) or \(k=4\). With odd \(k>1\), no solution.