Question

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.

Answer 1

Hint:
According to the question, you can check the degree of polynomial by applying the formula Dividend = Divisor x Quotient + Remainder . Hence, you can show that whether \[{x^2} - 1\] is satisfied or not.

Formula Used:
Here, we use the formula of division algorithm which is Dividend = Divisor x Quotient + Remainder

Complete step by step solution:
No, because whenever we divide a polynomial by a \[{x^6} + 2{x^3} + x - 1\] polynomial in x of degree 5, then we get quotient always as in a linear form i.e., polynomial in x of degree is 1. Let us assume, divisor = a polynomial in x of degree 5 which is equals to \[a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f\] quotient = \[{x^2} - 1\] and dividend = \[{x^6} + 2{x^3} + x - 1\]
So, by applying division algorithm for polynomials, we get
Dividend = Divisor x Quotient + Remainder
On putting all the values in above formula, we get
= \[a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f \times \] \[{x^2} - 1 + \] Remainder
Here, \[a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f \times \] \[{x^2} - 1\] is a polynomial of 7 degree.
So,
= (a polynomial of degree 7) + Remainder
[As we know, in division algorithm: degree of divisor > degree of remainder]= (a polynomial of degree 7)
But dividend = a polynomial of degree 6
Here, the degree of the product of the quotient and the divisor is 7. So, the degree of \[{x^6} + 2{x^3} + x - 1\] is 6
So, the division algorithm is not satisfied.

Hence, \[{x^2} - 1\] is not a required quotient.

Note:
To solve these types of questions, you can check by applying a division algorithm for polynomials. As well as you can check the degree of polynomial by putting all the given in the formula. Hence, you get the required answer.