2
$\begingroup$

Let's say we want to calculate the remainder of $$x^7+5x^3+2x+3 $$ divided by $$x^2 + 1$$

My teacher told us to this:

$$ x^7+5x^3+2x+3=(x^2+1)Q_{(x)}+r {(x)} $$ $$ ({x^2})^3x+5(x^2)x +2x+3=(x^2+1)Q_ {(x)}+r _{(x)}$$

Let $x^2=-1$

$$({-1})^3x+5(-1)x +2x+3=((-1)+1)Q_ {(x)}+r _{(x)}$$

So

$$r_{(x)}=-x-5x+2x+3=-4x+3$$

The only problem is: i don't understand why it's legal (!) to assign a value to $x^2$ and not $x $ (won't x be equal to $i$ anyway?) and then calculate the function.

(I know that If it was $x+1$ instead of $x^2+1$ i could calculate the function for $x=-1$ with no problem!)

Thanks for your explanation!

$\endgroup$
  • $\begingroup$ If you set $x^2=-1$ then you also must set $x=i$. $\endgroup$ – StubbornAtom Nov 18 '16 at 6:23
  • $\begingroup$ So is this method wrong? but it ends up with the right answer $\endgroup$ – Rima Nov 18 '16 at 6:25
  • $\begingroup$ Maybe the reason is that if you have $x=i$, it's like if you changed of variable your polynomial. $\endgroup$ – MonsieurGalois Nov 18 '16 at 6:30
1
$\begingroup$

The "trick" is legitimate, though maybe not entirely obvious.

Let $P(x)=x^7+5x^3+2x+3$, and define the polynomial in $2$ variables:

$$S(u,v)=v^3 u + 5 v u + 2 u + 3 $$

It is easily verified that $P(x)=S(x,x^2)$.

Consider now $S$ as a polynomial in $v$ alone with $u$ as a parameter, then the remainder of the division by $v+1$ will be $S(u,-1) = -4u+3$ so for some quotient $T(u,v)$:

$$S(u,v)=(v+1)T(u,v) + (-4u+3)$$

Now substitute $u=x, v=x^2$ back and you get:

$$P(x)=(x^2+1)Q(x) + (-4x+3)$$

where $Q(x)=T(x,x^2)$ so $\;-4x+3\;$ is indeed the remainder of the division of $P(x)$ by $x^2+1$.

$\endgroup$
  • $\begingroup$ So by assuming a two variable function: yes i can assign a value to any part(variable) of function i want. Is that correct? $\endgroup$ – Rima Nov 18 '16 at 7:30
  • 1
    $\begingroup$ @Rima At that point, think of $S$ as a polynomial in one variable, $v$. More formally, you can call that polynomial $S_u(v) = S(u,v)$. Then everything you know about polynomials in one single variable applies to $S_u(v)$ as well, including the remainder of the division by $v+1$ being $S_u(-1)$. $\endgroup$ – dxiv Nov 18 '16 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.