How do you long divide #6x^3 + 3x^2 - 4x + 7# by #x^2+1#?

1 Answer
Jun 15, 2015

#6x + 3 + (-10x +4)/(x^2 + 1)#

Explanation:

Let's write
#A(x)=6x^3 + 3x^2 - 4x + 7# and
#B(x)=x^2 + 1#

We use the rule: we find the first term dividing the term with the highest degree in #A(x)# by the term with the highest degree in #B(x)#, let's call it #q_(3-2)x^(3-2)# then we calculate #P_1(x)=A(x)-B(x)*Q_(3-2)(x)#, which has degree 2

#(6x^3)/x^2=6x => P_1(x)=6x^3 + 3x^2 - 4x + 7 -6x(x^2 + 1)=3x^2 -10x + 7#

Now we consider the same process with #P_1(x)#, and we have #q_(3-2-1)x^(3-2-1)# and #R(x)#.

#(3x^2)/x^2=3 => R(x)=3x^2 -10x + 7 - 3(x^2 + 1)=-10x +4#

Notice that

#del(R(x)) < del(B(x))#

so we're done, and #Q(x)=q_(3-2)x^(3-2) + q_(3-2-1)x^3-2-1 = 6x + 3# is the quotient and #R(x)# is the reminder

NB: I consider #del(P)# as the degree of #P#