P(x) = x^4 + x^2 + 2?

Question

P(x) = x^4 + x^2 + 2
d(x) = x^2 +x +1

what is the Q(x) and the R(x) ?

when P(x) is divided by d(x) and in the format P(x) = d(x) * Q(x) + R(x)

Answer 1

P(x) = x^4 + x^2 + 2, it is to be divided by d(x) = x^2 + x + 1

In P (x), Split 2 into (1+1) and let g(x) = x^4 + x^2 + 1

So P(x) = g(x) + 1

start with g(x) = x^4 + x^2 + 1 ------ (2)
Eq (2) is of the type (A^2 – B^2) like
(t^2 + t + 1) = (t^2 + t + 1 + t – t)
= (t^2 + 2t + 1 – t)
= {(t+1)^2 – t} = {(x^2+1)^2 – (x^2)}
= {(x^2+x+1) (x^2-x+1)}

g(x) = x^4 + x^2 + 1 = {(x^2+x+1) (x^2-x+1)} put in (1)

So P(x) = {(x^2+x+1)*(x^2-x+1)}+ 1 = d(x)* Q(x) + R(x)

So Q(x) = (x^2-x+1)} and R(x) = 1

Answer 2

Q(x) and R(x) are undefined until you define them.