A polynomial f(x) leaves remainder 2x + 1 when divided by 2x^2 - 3x + 4. What is the remainder when f(x)^3 is divided by 2x^2 - 3x + 4? Choose from 26x - 47, 27x - 48, 28x - 49, 29x - 50, or 30x - 51.

Step-by-step solution

1. 1

   Write what the given remainder means. There is a quotient $q(x)$ with

   $f(x)=q(x)\left(2x^{2}-3x+4\right)+(2x+1).$

   Write $D(x)=2x^{2}-3x+4$ for short.

2. 2

   Cube and discard everything divisible by D. Expanding $\left[q D+(2x+1)\right]^{3}$, every term except the last contains at least one factor of $D$, so it vanishes on division:

   $\{f(x)\}^{3}\equiv (2x+1)^{3}\pmod{D(x)}.$

   This is the key move — you never need to know $q(x)$ or even the degree of $f$.

3. 3

   Expand the cube.

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

4. 4

   Divide by D(x) — first step. The leading terms give $\dfrac{8x^{3}}{2x^{2}}=4x$:

   $8x^{3}+12x^{2}+6x+1-4x\left(2x^{2}-3x+4\right)=24x^{2}-10x+1.$

5. 5

   Divide by D(x) — second step. Now $\dfrac{24x^{2}}{2x^{2}}=12$:

   $24x^{2}-10x+1-12\left(2x^{2}-3x+4\right)=26x-47.$

   The degree has dropped below $2$, so the division stops here.

6. 6

   State and check the answer. The remainder is $26x-47$, which is option ①. Verifying: $(4x+12)\left(2x^{2}-3x+4\right)+26x-47=8x^{3}+12x^{2}+6x+1=(2x+1)^{3}$ ✓.