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Let $r(x)$ be the remainder when the polynomial $x^{135}+x^{125}-x^{115}+x^5+1$ is divided by $x^3-x$. Then

a. $r(x)$ is the zero polynomial
b. $r(x)$ is a nonzero constant
c. the degree of $r(x)$ is one
d. the degree of $r(x)$ is two

Would like some help solving this. How would I apply the remainder theorem?

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$P(x) = Q(x)(x^3-x) + R(x)$

Note that the degree of $R(x)$ is at most $2$, since it must be lower than the degree of $x^3-x$. Also, $x^3 - x = x(x^2 - 1) = x(x-1)(x+1)$.

So,

$P(x) = Q(x)\cdot x \cdot (x+1) \cdot (x-1) + R(x)$

Note that $P(0) = R(0) = 1, P(1) = R(1) = 3, P(-1) = R(-1) = -1$.

Immediately, choices a) and b) are excluded.

Choice c) is possible because the trend here does not exclude monotonic behaviour, so let's see if the the slope remains constant. The change in $R(x)$ as $x$ goes from $-1$ to $0$ is an increase of $2$, while the change in $R(x)$ as $x$ goes from $0$ to $1$ is again an increase of $2$. This implies a constant slope (gradient), and therefore choice c) is the correct answer.

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  • $\begingroup$ Easier: if we use modular arithmetic it can be done in 10 seconds of mental arithmetic, e.g. see my answer. $\endgroup$ – Bill Dubuque Oct 20 '18 at 18:05
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Hint:

Write $$p(x)=q(x)(x^3-x)+r(x)$$

Try to evalute $p(0)$ to obtain $r(0)$.

Also, try to evaluate $p$ at some other values that can help you to determine some values of $r$.

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$\begin{align} x(x^{\large 2}\!\!-\!1)\,\mid\, x(x^{\large 2n}\!-1)\!\\[.3em] \Rightarrow\ \bmod\ x^{\large 3}\!-\!x\!:\,\ x^{\large 1+2n}\equiv x\end{align}\quad\ $ so $\quad \begin{align}{\quad\, x^{\large 135} + x^{\large 125} - x^{\large 115} + x^{\large 5} + 1\\ \equiv\, x\ \ \, +\,\ \ x\ \ \,-\,\ \ x\,\ \ +\,\ \ x\ +\ 1\,\equiv\, 2x+1}\end{align}$

Remark $\ $ More generally, we can factor $x$ out of mods via the mod Distributive Law

$\begin{align}f\!-\!f_0\bmod xg = x((f-f_0)/x\bmod g)\, = &\,\ x(x^{\large 134}+x^{\large 124}-x^{\large 114}+x^{\large 4})\bmod x^{\large 2}-1\\ = &\,\ x(1\,\ \ +\,\ \ 1\ \ \,-\,\ \ 1\,\ \ +\,\ \ 1\,\ \ )\bmod \color{#c00}{x^2-1}\,=\, 2x \end{align}$

So $\,\ f-1 \equiv 2x\pmod{x(x^{\large 2}-1)},\ $ since $\,\ \color{#c00}{x^{\large 2}\equiv 1}\,\Rightarrow\,x^{\large 2n} = (\color{#c00}{x^{\large 2}})^{\large n}\equiv \color{#c00}1^{\large n}\equiv 1$

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