Question

Apply division algorithm to find the quotient q(x) and remainder r(x) in dividing f(x) by g(x) in each of the following:
$f(x)= x^{3} -6x^{2} + 11x -6,g(x)=x^{2}+x+1$

Answer 1

SOLUTION IS IN THE ATTACHMENT…  

DIVISION ALGORITHM for polynomials :  

If f(x) and g(x) are any two polynomials with g(x) ≠0 , then we can find polynomials q(x) and r(x) , such that  

f(x) = g(x)  × q(x) + r(x)

Dividend =  Divisor ×  Quotient + Remainder

Where, r(x) = 0 or degree of r(x) < degree of g(x)  

This result is known as a division algorithm for polynomials.

★★ If r(x) = 0, then  polynomials g(x) is a factor of polynomial f(x) .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

how to divide with remainder by any monic polynomial, i.e any polynomial ffwhose leading coefficient c=1c=1 (or a unit), since ff monic implies that the leading term of ff divides all higher degree monomials xk, k≥n=deg f,xk, k≥n=deg f,so the division algorithm works to kill all higher degree terms in the dividend, leaving a remainder of degree <n=deg f.<n=deg f.

But this generally fails if ff is not monic, e.g. x=2xq+rx=2xq+r has no solution for r∈Z, q∈Z[x],r∈Z, q∈Z[x], since evaluating at x=0x=0 ⇒⇒ r=0,r=0, evaluating at x=1x=1 ⇒⇒ 2|12|1 in ZZ ⇒⇐⇒⇐ Notice that the same proof works in any coefficient ring RR in which 22 is not a unit (invertible). Conversely, if 22 is a unit in R,R, say 2u=12u=1 for u∈R,u∈R, then division is possible: x=2x⋅u+0.x=2x⋅u+0.

However, it is possible to divide with remainder by non-monic polynomials as follows.

Theorem (nonmonic Polynomial Division Algorithm)    Let 0≠F,G∈A[x]0≠F,G∈A[x] be polynomials over a commutative ring A,A,with aa = lead coef of F,F, and i≥max{0,1+degG−degF}.i≥max{0,1+deg⁡G−deg⁡F}.Then
aiG=QF+R  for some  Q,R∈A[x], degR<degF