\(x^4+4x^3+2x^2+x+4 = Q(x)\cdot (x^2+3x)+ax+b,\;we\;need\;to\;find\;a,\;and\;b;\\\\x^4+4x^3+2x^2+x+4 = Q(x)\cdot x\cdot (x+3)+ax+b\)
If you choose x = 0, then we have 4 = Q(0)0 + a0 + b, therefore b = 4.
If we choose x = -3, then we have:
\((-3)^4+4\cdot (-3)^3+2 \cdot (-3)^2-3+4 = Q(-3)\cdot (-3)\cdot (-3+3)-3a+4\Rightarrow 81-108+18+1=4-3a,\;or\;-12=-3a,\;so\;a=4.\)
The remainder is: 4x+4.
Green eyes.
