3

A monic cubic $p(x) $ is divided by $(x^2 +x +1)$ and the remainder is $(2x+3)$. When $p(x)$ is divided by $x(x+3)$, the remainder is $5(x+1)$. Find $p(x)$

So the way I solved it was through using long division - as you can imagine very lengthy, and not very efficient in exam situations. I then equated the remainders with the given remainders. However, in an exam situation, would there be an even faster way? I am open to suggestions.

Albus Dumbledore
  • 11,860
user71207
  • 1,543
  • isnt the degree of p(x) given? – Albus Dumbledore Nov 29 '20 at 06:19
  • 1
    sorry , i forgot to write cubic – user71207 Nov 29 '20 at 06:26
  • 3
    I suspect that long division is the intended solution. Typically, in an exam or homework setting, you are not expected to invent a new approach. Instead, the problem(s) are usually designed to reinforce the concepts that you've been taught. Therefore, if you hadn't already been taught an alternative approach, the problem is probably designed to test your inherent understanding of polynomial long division. Therefore, the approach that you took was probably the intended approach. – user2661923 Nov 29 '20 at 06:32
  • but just for practice at home, I wonder if there is a faster method to save time – user71207 Nov 29 '20 at 06:34

2 Answers2

5

here is a trick: as the polynomial has leading coefficient 1 and degree 3 we have $$p(x)=(x+K)(x^2+x+1)+2x+3$$ Now use the fact that $$p(x)=x(x+3)g(x)+5(x+1)$$ $$p(0)=5(0+1)=5$$ $$K=2$$

Albus Dumbledore
  • 11,860
  • 1
    Magnificent. Such a concise solution. However, this method seems to only work given a specific set of conditions right? ie it will not work all the time – user71207 Nov 29 '20 at 06:46
  • @user71207 maybe not ,sometimes other tricks may work,this is is good as long as the degree of polynomial is close to the one dividing – Albus Dumbledore Nov 29 '20 at 06:48
  • 1
    (+1) This is certainly quite simple. It is tailored to this particular case, but that is something shared by most short-cuts. – robjohn Nov 29 '20 at 17:41
1

This may not be a quicker way, but it is systematic and will work in all cases where the two moduli have no common factors. Hopefully, it also might make the use of the Extended Euclidean Algorithm with the Chinese Remainder Theorem more transparent.

We want $$ \begin{align} p(x)&\equiv2x+3&&\pmod{x^2+x+1}\tag{1a}\\ p(x)&\equiv5x+5&&\pmod{x^2+3x}\tag{1b} \end{align} $$ so we apply the Extended Euclidean Algorithm (as implemented in this answer and adapted to polynomials) to $x^2+3x$ and $x^2+x+1$: $$ \begin{array}{c|c|c|c} x^2+3x&1&0\\ x^2+x+1&0&1\\ 2x-1&1&-1&1\\ \frac74&-\frac{2x+3}4&\frac{2x+7}4&\frac{2x+3}4 \end{array}\tag2 $$ Thus, we get $$ (2x+7)\left(x^2+x+1\right)-(2x+3)\left(x^2+3x\right)=7\tag3 $$ So we have $$ -\frac{(2x+3)\left(x^2+3x\right)}7\equiv\left\{ \begin{align} &1&&\pmod{x^2+x+1}\\ &0&&\pmod{x^2+3x} \end{align} \right.\tag4 $$ and $$ \frac{(2x+7)\left(x^2+x+1\right)}7\equiv\left\{ \begin{align} &0&&\pmod{x^2+x+1}\\ &1&&\pmod{x^2+3x}\\ \end{align} \right.\tag5 $$ Adding $2x+3$ times $(4)$ to $5x+5$ times $(5)$ yields one solution to $(1)$: $$ \begin{align} &\overbrace{(5x+5)\frac{(2x+7)\left(x^2+x+1\right)}7}^{\substack{0&\pmod{x^2+x+1}\\5x+5&\pmod{x^2+3x}}}\overbrace{-(2x+3)\frac{(2x+3)\left(x^2+3x\right)}7}^{\substack{2x+3&\pmod{x^2+x+1}\\0&\pmod{x^2+3x}}}\\ &=\frac{6x^4+31x^3+45x^2+53x+35}7\tag6 \end{align} $$ The Chinese Remainder Theorem says that this is the solution mod $\left(x^2+x+1\right)\left(x^2+3x\right)$, so we can subtract $\frac67$ of the modulus to get $$ \begin{align} p(x) &=\frac{6x^4+31x^3+45x^2+53x+35}7-\frac67\left(x^2+x+1\right)\left(x^2+3x\right)\\ &=\bbox[5px,border:2px solid #C0A000]{x^3+3x^2+5x+5}\tag7 \end{align} $$

robjohn
  • 345,667