Find the quotient and the remainder of $(n^6-7)/(n^2+1)$

Question

Given that $n$ belong to $\mathbb{N}$.

Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.

So I tried to divide them up and got a negative expression $(-n^4-7)$.

How to continue?

Or what can be done differently?

How to find the quotent and the remainder?

Answer 1

I assume that by "dish" you mean "quotient"; that's not terminology I'm familiar with, but it seems to make sense in context.

When you're dividing polynomials, remember that you are not done until the remainder is of lower degree than the denominator. So you're correct that the first step of dividing $n^2 + 1$ into $n^6 - 7$ gives you a remainder of $-n^4 - 7$ (with $n^4$ in the quotient); but since $-n^4 - 7$ is of higher degree than $n^2 + 1$, we have to keep going.

To go into $-n^4 - 7$, we need to multiply by $-n^2$; so now our quotient so far is $n^4 - n^2$, and our remainder is $(-n^4 - 7) - (-n^2 (n^2 + 1)) = n^2 - 7$.

Since $n^2 - 7$ is still at least as high in degree as $n^2 + 1$, we have to divide one more time; this gives us a quotient of $n^4 - n^2 + 1$ and a remainder of $(n^2 - 7) - (n^2 + 1) = -8$.

Since the remainder now has degree $0$, which is less than the degree of our divisor, we can stop now; so the quotient is $n^4 - n^2 + 1$ and the remainder is $-8$.

Answer 2

We have $n^6-7 = (n^2+1)q(n)+an+b$, with $a,b \in \mathbb Z$, because $n^2+1$ is monic.

Plugging $n=\pm i$, we get $$ -8 = \pm ai+b $$ which implies $a=0$ and $b=-8$. Therefore, $$ q(n) = \frac{n^6-7+8}{n^2+1} = \frac{n^6+1}{n^2+1} = n^4 - n^2+ 1 $$ Thus, $$ n^6-7 = (n^2+1)(n^4 - n^2+ 1)-8 $$

Answer 3

$$ \left( x^{6} - 7 \right) $$

$$ \left( x^{2} + 1 \right) $$

$$ \left( x^{6} - 7 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x^{4} - x^{2} + 1 \right) } + \left( -8 \right) $$ $$ \left( x^{2} + 1 \right) = \left( -8 \right) \cdot \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x^{4} - x^{2} + 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x^{4} - x^{2} + 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{6} + 7 }{ 8 } \right) }{ \left( \frac{ - x^{2} - 1 }{ 8 } \right) } $$ $$ \left( x^{6} - 7 \right) \left( \frac{ 1}{8 } \right) - \left( x^{2} + 1 \right) \left( \frac{ x^{4} - x^{2} + 1 }{ 8 } \right) = \left( -1 \right) $$

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Answer 4

I was going to answer your question exactly in the way lhf have done, just explaining the process in my post. Firstly, if we divide a polynomial $f(n)$ by another polynomial $g(n)$ of lesser degree, we will have a remainder $h(n)$ such that, $\text{deg(}h)< \text{deg(g)}$. $$f(n)=g(n)\cdot q(n)+h(n)~~~~\cdots (1) $$ as, if you have term with power higher or equal to $\text{deg(}g)$ then, you can fit it in the polynomial $q(n)$, so, you will always get such an unique $h(n)$, and we call this $h(n)$ as remainder polynomial.

Now, if we divide the given polynomial $n^6-7$ by $n^2+1 $ then we have remainder with degree atmost $1$(as, $\text{deg(}n^2+1)=2$. Hence, it will be of the form $an+b$. $$n^6-7 = (n^2+1)\cdot q(n)+an+b~~~~~~~\cdots (2)$$

Now, the only information that will help us to find the remainder are the roots of $(n^2+1)$, $n=\pm i$, as we don't know what is $q(n)$, the roots will help to vanish the first term of equation $(2)$. The first term will be zero $($we are choosing the roots of $n^2+1$$)$ and we will left with second term only of equation $(2)$ and we will get the following equations: $$ai+b=-8\\-ai+b=-8$$ which gives, $a=0,b=-8$. So, $$n^6-7=(n^2+1)\cdot q(n)-8\implies n^6-7+8=(n^2+1)\cdot q(n)$$ factorizing we get, $n^6+1=(n^2)^3+1=(n^2+1)(n^4-n^2+1)$. Hence, $q(n)= n^4-n^2+1$. Finally we get, $$n^6-7=(n^2+1)\cdot (n^4-n^2+1)-8$$

Answer 5

Yet another way: $$ \begin{align} n^6-7 &= \big(\left(n^2\color{red}{+1}\right)\color{red}{-1}\big)^3-7 \\ &= \left(n^2+1\right)^3-3\left(n^2+1\right)^2+3\left(n^2+1\right)-1 -7 \\ &= \left(n^2+1\right)\big(\left(n^2+1\right)^2-3\left(n^2+1\right)+3\big) - 8 \\ &= \left(n^2+1\right)\left(n^4-n^2+1\right) - 8 \end{align} $$

Answer 6

$\bmod n^2\!+1\!:\,\ \color{#c00}{n^2\equiv -1}\,\Rightarrow\, n^6\equiv (\color{#c00}{n^2})^3\equiv (\color{#c00}{-1})^3\equiv -1$ so $\,n^6-7\equiv -8$ is the remainder,

hence the quotient is $\ \dfrac{(n^6-7)-8}{n^2+1} = \dfrac{n^6+1}{n^2+1} = n^4 - n^2 + 1$