(i) No, because whenever we divide a polynomial x⁶+2x³+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
=ax⁵+bx⁴+cx³+dx²+ex+f
quotient =x²-1
and dividend =x⁶+2x³+x-1
By division algorithm for polynomials,
Dividend = Divisor × Quotient + Remainder
=(ax⁵+bx⁴+cx³+dx²+ex+f)×(x2-1)+ Remainder
=(a polynomial of degree 7)+ Remainder
[in divisor algorithm, degree of divisor > degree of remainder]
=(a polynomial of degree7)
But divident = a polynomial of degree 6
So, division algorithm is not satisfied.
Hence, x²-1 is not a required quotient.
(ii) Given that, Divisor px³+qx²+rx+s,p≠0
and dividend =ax²+bx+x
We see that, Degree of divisor > Degree of dividend
So, by divison algorithm,
quotient = 0 and remainder =ax²+bx+c
If degree of dividend < 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)< 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)≤ degree of p(x).
(v) No, let p(x)=x²+kx+k
If p(x) has equal zeroes, then its discriminant should be zero.
∴D=B²-4AC=0 …(i)
On comparing p(x) with Ax²+Bx+C, we get
A=1,B=k and c=k
∴(k)²-4(1)(k)=0 [from Eq.(i)]
⇒k(k-4)=0
→k=0,4
So, the quadratic polynomial p(x) have equal zeroes only at k=0,4.