# Why do we need to consider multiplication of negative numbers in polynomial equations to arrive at all solutions?

I'm a computer science graduate and I'm not sure to which level of Algebra this question belongs to, whether it is abstract algebra or some other filed, but I'm curious to know and learn.

First, before the usage of negative numbers in multiplication, multiplication was defined as repeated addition of positive numbers(or integers), so we can have the equation $$x^2 -4x + 3 = 0$$

Where I mean by the multiplication in $$x^2$$ and $$4x$$ multiplication as repeated addition, I know nothing about negative numbers multiplication yet.

We can rearrange the terms and factorize so we have:

$$(x-2)(x-2) = 1$$

Now, if we think of multiplication as repeated addition, with no mention of negative numbers multiplication, then $$(x-2)$$ can only be positive, so we have $$(x-2) ^2 = 1$$ so we have one solution, namely, $$x = 3$$.

But for me, here comes the problem: there is another solution, namley $$x = 1$$, and we can find this solution only in case we allowed $$(x-2)$$ to be negative number, and defined multiplication of negative numbers by negative numbers so for example

$$-1 * -1 = 1$$

in which case $$(x-2) = - 1$$ and we have $$x = 1$$.

So how come that we first dealt with multiplication as repeated addition, but then we needed to define(or extend) it in another way so that we can get all the solution of an equation that contained terms multiplied in the sense of repeated addition only?

Can someone recommend a textbook that explain this and related stuff from the beginning, illustrating the motivation and building on this possibly also for complex numbers as well? Again I'm not sure in which branch or level of Algebra these stuff are explained.

Edit:

By the way my question isn't: "why is multiplication of negative numbers defined as it is", which answer is: "to preserve the structure", but rather why must we consider multiplication of negative numbers in the mentioned example equation to arrive at all solutions ! That is it seems that multiplication using negative numbers in the mentioned equation is so inherent and fundamental. I'm not sure if I phrased my question correctly but here it goes...

• See math.stackexchange.com/questions/156264/…, math.stackexchange.com/questions/9933/…, and many other questions. Mar 13 at 18:24
• Note that we can also find the solution $x=1$ by rearranging the equation to have no instances of subtraction, namely as $$x^2+3=4x.$$ Now verifying that $x=1$ is a solution is entirely within the "positive realm." You should think of the extension of multiplication to negative numbers as being governed by the desire to stay true to algebraic facts we already know. Mar 13 at 18:42
• That's simple: completing the square (quadratic formula) transforms to a reduced quadratic $\,\bar x^2 = d\,$ which has a positive and negative root (when $d>0)$. So even if the original roots $\,x\,$ are both $> 0\,$ the arithmetic takes a detour through negative numbers. Similarly the computation may involve fractions even if the roots are integers, and the cubic formula may take a detour through complex numbers even for real roots. Mar 14 at 9:13
• Only by passing to these number systems extending $\Bbb N$ can we obtain a single uniform formula for solving quadratics. In ancient times before negatives, fractions or complexes were invented, they had no such quadratic formula - instead they had to consider numerous cases to ensure they worked only with positive integers. Enabling uniform methods for solving polynomial equations is one of the primary motivations that led to such extended number systems. Mar 14 at 9:13
• Well, you could've factorised it as $(2-x)(2-x)=1$ also, at least in this particular example... Mar 15 at 18:11