# Answer the following and justify:Can $x^{2}-1$ be the quotient on division of $x^{6}+2 x^{3}+x-1$ by a polynomial in $x$ of degree 5 ?

To do:

We have to answer the given questions and justify them.

Solution:

(i) Let the divisor, a polynomial in $x$ of degree 5 be $ax^5 + bx^4 + cx^3 + dx^2 + ex + f$

Quotient $= x^2 -1$

By division algorithm for polynomials,

Dividend $=$ Divisor $\times$ Quotient $+$ Remainder

$= (ax^5 + bx^4 + cx^3 + dx^2 + ex + f)\times(x^2 -1) +$ Remainder

$=$ (a polynomial of degree 7) $+$ Remainder

But the given dividend is a polynomial of degree 6.

Here, the division algorithm is not satisfied.

Hence, $x^2 -1$ cannot be the quotient on division of $x^{6}+2 x^{3}+x-1$ by a polynomial in $x$ of degree 5.

(ii) Here,

Divisor $=px3 + qx2 + rx + s, p≠0$

Dividend $= ax^2 + bx + c$

We observe that,

Degree of divisor $>$ Degree of dividend

We know that,

If the degree of dividend is less than the degree of the divisor, then the quotient will be zero and the remainder is same as the dividend.

Therefore, by division algorithm,

Quotient $= 0$ and Remainder $= ax^2 + bx + c$

(iii) If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, then the relation between the degrees of p(x) and g(x) is the degree of p(x) is less than the degree of g(x).

For example,

$p(x)=10x$ and $g(x)=5x^2$

(iv) If on 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 a degree less than or equal to the degree of p(x).

For example,

$p(x)=10x^2$ and $g(x)=5x$ then $p(x) \div g(x)=10x^2 \div 5x=2x$

$p(x)=10x^2$ and $g(x)=5x^2$ then $p(x) \div g(x)=10x^2 \div 5x^2=2$

(v) Let $p(x) = x^2 + kx + k$

If $p(x)$ has equal zeroes, then its discriminant is zero.

$D = b^2 -4ac = 0$                        Here,

$a =1, b = k$ and $c = k$

Therefore,

$(k)^2-4(1)(k) = 0$

$k(k- 4)=0$

$k =0$ or $k=4$

This implies, the quadratic polynomial $p(x)$ has equal zeroes at $k =0, 4$.

Hence, the quadratic polynomial $x^{2}+k x+k$ cannot have equal zeroes for some odd integer $k>1$.