A polynomial f(x) leaves a remainder of 2x + 1 when it is divided by 2x^2 - 3x + 4. Find the remainder when [f(x)]^3 is divided by 2x^2 - 3x + 4.

Step-by-step solution

1. 1

   Turn the division fact into an equation you can cube. "Remainder $2x+1$" means there is some quotient polynomial $Q(x)$ with

   $f(x)=Q(x)\,(2x^2-3x+4)+(2x+1)$

   This is the only information given, so every step has to flow from it. Write $D(x)=2x^2-3x+4$ for short.

2. 2

   Cube both sides and throw away everything divisible by $D(x)$. Expanding $\big(QD+(2x+1)\big)^3$ with the binomial theorem gives

   $Q^3D^3+3Q^2D^2(2x+1)+3QD(2x+1)^2+(2x+1)^3$

   The first three terms all contain a factor of $D(x)$, so they contribute nothing to the remainder. That is the key idea: the remainder of $f^3$ equals the remainder of $(2x+1)^3$. You never need to know $f$ or $Q$ at all.

3. 3

   Expand the cube.

   $(2x+1)^3=8x^3+12x^2+6x+1$

   The divisor has degree $2$, so the remainder we are hunting for has degree at most $1$ — it will look like $Ax+B$.

4. 4

   Reduce the powers of $x$ using the divisor itself. Instead of doing long division, use the relation $2x^2-3x+4=0$ in the quotient ring, i.e. treat

   $x^2=\frac{3x-4}{2}$

   as a rewriting rule. Then

   $x^3=x\cdot x^2=\frac{3x^2-4x}{2}=\frac{3\cdot\frac{3x-4}{2}-4x}{2}=\frac{\frac{9x-12-8x}{2}}{2}=\frac{x-12}{4}$

   This substitution trick is much faster than dividing a cubic by a quadratic, and it generalises to any power of $x$.

5. 5

   Substitute and collect like terms.

   $8x^3=8\cdot\frac{x-12}{4}=2x-24,\qquad 12x^2=12\cdot\frac{3x-4}{2}=18x-24$

   so

   $(2x+1)^3\equiv(2x-24)+(18x-24)+6x+1=26x-47$

6. 6

   Check with an honest long division. Dividing $8x^3+12x^2+6x+1$ by $2x^2-3x+4$ gives quotient $4x+12$, and

   $(2x^2-3x+4)(4x+12)=8x^3+12x^2-20x+48$

   Adding the claimed remainder: $8x^3+12x^2-20x+48+26x-47=8x^3+12x^2+6x+1$, which is exactly the cube. The remainder is confirmed.