Exercise 1
Answer the following and justify:
(i)
Can be the quotient on division of by a polynomial in of degree ?
(ii)
What will the quotient and remainder be on division of by ?
(iii)
If on division of a polynomial by a polynomial , the quotient is zero, what is the relation between the degrees of and ?
(iv)
If on division of a non-zero polynomial by a polynomial , the remainder is zero, what is the relation between the degrees of and ?
(v)
Can the quadratic polynomial have equal zeroes for some odd integer ?
Matches the book, not yet reviewed
This working reaches the answer NCERT prints. It has not yet been read through by hand.
NCERT’s answer
(i)
No
(ii)
\(\displaystyle 0, a x^2+b x+c\)
(iii)
\(\displaystyle \operatorname{deg} p(x)<\operatorname{deg} g(x)\)
(iv)
\(\displaystyle \operatorname{deg} g(x) \leq \operatorname{deg} p(x)\)
(v)
No
(i)
No — \[\deg p=\deg g+\deg q \] forces \(\displaystyle \deg q=6-5=1\), but \(\displaystyle x^2-1\) has degree \(\displaystyle 2\).
(ii)
Quotient \(\displaystyle 0\), remainder \(\displaystyle ax^2+bx+c\) — \[\deg(ax^2+bx+c)<\deg(px^3+qx^2+rx+s) \] so division stops immediately.
(iii)
\[p(x)=g(x)\cdot 0+r(x)=r(x),\quad \deg r(x)<\deg g(x) \Rightarrow \deg p(x)<\deg g(x) \]
(iv)
\[p(x)=g(x)q(x),\ q\neq 0 \Rightarrow \deg p(x)=\deg g(x)+\deg q(x)\ge \deg g(x) \]
(v)
No — \[k^2-4k=0 \Rightarrow k=0,\,4 \] neither is an odd integer greater than \(\displaystyle 1\).
Answer: (i) No (ii) quotient \(\displaystyle 0\), remainder \(\displaystyle ax^2+bx+c\) (iii) \(\displaystyle \deg p(x)<\deg g(x)\) (iv) \(\displaystyle \deg p(x)\ge \deg g(x)\) (v) No.