(i) No, because whenever we divide a polynomial `x^(6) +2x^(3) +x -1` by a polynomial in x of degree 5, then we get quotient always as in linear form i.e., polynomial in x of degree 1.
Let divisor = a polynomial in x of degree 5
`=ax^(5) +bx^(4) +cx^(3) +dx^(2) +ex +f`
quotient `= x^(2) - 1`
and dividend `= x^(6) + 2x^(3) +x - 1`
By division algorithm for polynomials,
Dividend = Divisor `xx` Quotient + Remainder
`= (ax^(5)+bx^(4)+c^(3)+dx^(2)+ex +f) xx (x^(2)-1)+` Remainder
=(a polynomial of degree 7)+ Remainder
[in divisor algorithm, degree of divisor `gt` degree of remainder]
=(a polynomial of degree7)
But divident = a polynomial of degree 6
So, division algorithm is not satisfied.
Hence, `x^(2) -1` is not a required quotient.
(ii) Given that, Divisor `px^(3) +qx^(2) +rx + s, p ne 0`
and dividend `= ax^(2) +bx +x`
We see that, Degree of divisor `gt` Degree of dividend
So, by divison algorithm,
quotient = 0 and remainder `= ax^(2) +bx +c`
If degree of dividend `lt` degree of divisor, then quotient will be zero and remainder as same as dividend.
(iii) If division of a polynomial `p(x)` by a polynomial `g(x)`, the quotient is zero, then relation between the degrees of `p(x)` and `g(x)` is degree of `p(x) lt` degree of `g(x)`.
(iv) If division of a non-zero polynomial `p(x)` by a polynomial `g(x)`, the remainder is zero, then `g(x)` is a factor of `p(x)` and has degree less than or equal to the degree of `p(x)`. i.e., degree of `g(x) le` degree of `p(x)`.
(v) No, let `p(x) = a^(2) +kx +k`
If `p(x)` has equal zeroes, then its discriminant should be zero.
`:. D = B^(2)-4AC = 0` ...(i)
On comparing `p(x)` with `Ax^(2)+Bx +C`, we get
`A = 1, B = k` and `c = k`
`:. (k)^(2)-4(1)(k)=0` [from Eq.(i)]
`rArr k(k-4) = 0`
`rarr k = 0, 4`
So, the quadratic polynomial `p(x)` have equal zeroess only at `k = 0,4`.
