Question

Question 1 (i)
Answer the following and justify
Can $x^2–1$ be the quotient on division of $x6+2x^3+x-1$ by a polynomial in x of degree 5?

Answer 1

(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 division = a polynomial in x of degree 5
$=a{x}^5+b{x}^3+c{x}^3+d{x}^2+ex+1$
Quotient = $x^2-1$
And dividend = $x^6+2x^3+x–1$
By division algorithm for polynomials
Dividend = divisor $\times$ quotient + remainder
$=(a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+f)\times (x^2-1)$+ remainder
= (a polynomial of degree 7) + remainder
[in division algorithm, degree of divisor > degree of remainder]
= (a polynomial of degree 7)
But dividend = a polynomial of degree 6
So, division algorithm is not satisfied
Hence, x2 – 1 is not a required quotient.