f(x) is a cubic polynomial and (x + 1)f(x) leaves remainder x + 2 when divided by x^3 + 2. If f(0) = 4, find f(2).

Step-by-step solution

1. 1

   Write down the division algorithm. "Remainder $x+2$ on division by $x^3+2$" means there is a polynomial $Q(x)$ with

   $(x+1)f(x)=(x^3+2)Q(x)+x+2$

   This single identity is the whole content of the hypothesis, so everything else follows from it.

2. 2

   Pin down the degree of $Q$. Since $f$ is cubic, $(x+1)f(x)$ has degree $4$. The product $(x^3+2)Q(x)$ must supply that degree $4$, so $\deg Q=1$. Write

   $Q(x)=ax+b$

   This is why the problem is solvable at all: only two unknowns remain.

3. 3

   Use the factor $x+1$ on the left. The left side vanishes at $x=-1$, so the right side must too. Substituting $x=-1$ into $(x^3+2)(ax+b)+x+2$:

   $(-1+2)(-a+b)+1=0\;\Longrightarrow\;b-a+1=0\;\Longrightarrow\;b=a-1$

4. 4

   Use $f(0)=4$. Substituting $x=0$ gives $(0+1)f(0)=4$ on the left and $(0+2)(b)+0+2=2b+2$ on the right, so

   $2b+2=4\;\Longrightarrow\;b=1\;\Longrightarrow\;a=2$

   Hence $Q(x)=2x+1$.

5. 5

   Evaluate at $x=2$. With $Q$ known,

   $(2+1)f(2)=(2^3+2)(2\cdot 2+1)+2+2=10\cdot 5+4=54$

   so $3f(2)=54$ and $f(2)=18$.

6. 6

   Check by recovering $f$ explicitly. Expanding, $(x+1)f(x)=2x^4+x^3+5x+4$. Dividing by $x+1$ gives

   $f(x)=2x^3-x^2+x+4$

   Indeed $f(0)=4$ and $f(2)=16-4+2+4=18$, and $2x^4+x^3+5x+4-(x^3+2)(2x+1)=x+2$ exactly.