Why is performing synthetic division of polynomials acceptable when one may be zero? ${}$

Question

So there is a typical problem in quadratic equations. Consider one as an example:

If $x=3 +\sqrt 5$ then find $x^4 -12x^3 +44x^2 -48x +17$.

The solution is pretty simple and a similar approach is put forward in another problem in complex numbers. Certainly we proceed by squaring $x=3 +\sqrt 5$ to reach $x^2 -6x +4 =0$.

Then by Synthetic Division/Division Algorithm we reach as follows: $x^4 -12x^3 +44x^2 -48x +17 = (x^2 -6x +4)(x^2 -6x +4) + 1$.

But since $x^2 -6x +4 = 0$ then the above expression reduces to 1, that’s the answer.

But we did remembered the above formula as Dividend = Divisor $\times$ Quotient + Remainder. Now the Divisor ($x^2 -6x +4$ in this case) ${} = 0$, then how come we can divide something by zero ($0$)? Pardon me for any mistake I am just first question new to stack exchange world. Please do tell me how do we insert mathematical symbols in the question.

Thank You

Answer 1

Let $f(x) = x^4 -12*x^3 +44*x^2 -48*x +17.$

Let $\displaystyle g(x) = x- \left(3 +\sqrt{5}\right).$

Let $h(x) = [g(x)]^2.$

When you divide $f(x)$ by $h(x)$, you are not evaluating the function $f(x)$ at any particular value of $(x)$, or evaluating $h(x)$ at any particular value of $(x)$.

Instead, you are treating $f(x)$ and $h(x)$ as polynomial functions, that are each members of the field of such polynomial functions (of finite degree), where each coefficient of the corresponding function is a Real number.

This is totally different than performing the evaluation of one or more of the polynomial functions at specific values of $(x)$, and then performing a division, based on such an evaluation.

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As a much simpler example,

let $j(x) = x-1$ and let $k(x) = x^2 - 3x + 2.$

Then $j(x)$ is a factor of $k(x)$ even though $j(1) = 0 = k(1).$

Answer 2

You are doing polynomial long division, and while the divisor quadratic may be zero at $x=3+\sqrt5$, it is not zero in general, so you need not worry about division by zero.

Polynomial long division only performs actual, arithmetic divisions in the field of coefficients, not the ring of polynomials. The successive reductions of degree of the dividend are a sequence of multiplications by powers of $x$ and subtractions, not involving divisions. (Hence the alternative name "Euclidean division".)

Answer 3

There is a set of algebraic fractions called $\Bbb R(x)$ which, roughly speaking, is the set of quotients of polynomials $\dfrac{P(x)}{Q(x)}$ where we consider $\dfrac{P(x)}{Q(x)}=\dfrac{P'(x)}{Q'(x)}$ when $P(x)Q'(x)=P'(x)Q(x)$. (The symbol $'$ is has nothing to do with derivatives here).

In this set $x$ is not a number, just a symbol. It is possible that $Q$ divides $P$, and in this case $\dfrac{P(x)/Q(x)}{1}=\dfrac{P(x)}{Q(x)}$, just like we consider the fraction $\dfrac 62$ equal to the integer number $3$.

On the other side, we have rational functions, that are quotients of polynomial functions.

For example, the function $$f(x)=\dfrac{x^2-1}{x-1}\text{ for }x\neq 1$$ is a rational function that is it not defined ad $x=1$ because the denominator becomes $0$ for that value. And the function $$\tilde f(x)=x+1\text{ for }x\in\Bbb R$$ is a polynomial function and its domain is $\Bbb R$.

Since the expressions of the functions, as algebraic fractions, are equivalent, $f(a)=\tilde f(a)$ for $a\neq 1$, but for $a=1$, $f(1)$ is not defined but $\tilde f(1)=2$

To sum up, as algebraic fractions $\dfrac{x^2-1}{x-1}$ and $x+1$ are the same, but as functions, $f(x)=\dfrac{x^2-1}{x-1}$ and $\tilde f(x)=x+1$, are not.

Answer 4

You do the operations without using the word "division". Instead use the words "substitution" and "elimination".

Having found that $x^2-6x+4=0$,rearrange this to give

$x^2=6x-4.$

Now in your polynomial function

$f(x)=x^4-12x^3+44x^2-48x+17,$

you use the quadratic equation to render

$x^4=6x^3-4x^2$

and then eliminate the $x^4$ term from $f(x)$:

$f(x)=(6x^3-4x^2)-12x^3+44x^2-48x+17=-6x^3+40x^2-48x+17.$

In subsequent steps we use the relations

$x^3=6x^2-4x$

$x^2=6x-4$

to eliminate the cubic and then the quadratic terms in the same way. Thus without using any division operation and therefore without any issue with division by zero. We reduce $f(x)$ to $Ax+B$ for some constants $A$ and $B$. In this case $A=0,B=1$ so we get $1$ as our final result; but if $A$ were not zero the evaluation still would have been much simpler that directly substituting into the fourth-degree polynomial.