Question

Let $\mathit{f}\left(x\right)\mathbf{,}\mathit{g}\left(x\right)\mathbf{,}\mathit{p}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$, with p (x) nonzero. Determine whether $\mathit{f}\left(x\right)\mathbf{\equiv }\mathit{g}\left(x\right)\left(modp\left(x\right)\right)$. Show your work.

1. $\mathit{f}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{5}}\mathbf{-}\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{4}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{g}\left(x\right)\mathbf{=}\mathbf{3}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}\mathbf{;}\mathbf{}\mathit{p}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{F}\mathbf{=}\mathbf{\mathbb{Q}}$
2. $\mathit{f}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{g}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{p}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{;}\mathbf{}\mathit{F}\mathbf{=}{\mathbf{\mathbb{Z}}}_{\mathbf{2}}$
3. role="math" localid="1649049604505" $\mathit{f}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{5}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{4}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{g}\left(x\right)\mathbf{=}\mathbf{3}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{9}\mathbf{;}\mathbf{}\mathit{p}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{;}\mathbf{}\mathit{F}\mathbf{=}\mathbf{\mathbb{Q}}$
   

Short answer

1. Yes, $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.
2. Yes, $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.
3. No,$f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.

Step-by-step solution

Part (a)

To determine whether $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$, find f (x) - g (x).

$$\begin{gathered} f\left(x\right)-g\left(x\right)=\left(x^5-2x^4+4x^3+x+1\right)-\left(3x^4+2x^3-5x^2-9\right) \\ =x^5-5x^4+2x^3+5x^2+x+10 \end{gathered}$$

Next, divide the obtained expression by x² + 1 as follows:

As the remainder is 0, so $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.

Part (b)

To determine whether $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$, find f (x) - g (x).

$$\begin{gathered} f\left(x\right)-g\left(x\right)=\left(x^4+x^3+x^2+1\right)-\left(x^4+x^2+x+1\right) \\ =x^3+x \end{gathered}$$

Next, divide the obtained expression by x² + 1 as follows:

Remainder will be 0, because $F={\mathrm{\mathbb{Z}}}_2$.

As the remainder is 0, $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.

Part (c)

To determine whether $f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$, find f (x) - g (x).

$$\begin{gathered} f\left(x\right)-g\left(x\right)=\left(3x^5+4x^4+5x^3-6x^2+5x-7\right)-\left(2x^5+6x^4+x^3+2x^2+2x-5\right) \\ =x^5-2x^4+4x^3-8x^2+3x-2 \end{gathered}$$

Next, divide the obtained expression by x² + 1 as follows:

As the remainder is not 0,$f\left(x\right)\equiv g\left(x\right)\left(\mathrm{mod}p\left(x\right)\right)$.