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