Intuition on why does $P(-1) = 0$ leads to identify one factor of a 3rd Degree Polynomial?

Question

Given $P(x) = x^3 + 3x^2 -13x -15$

In order to detect occasions in which $P(x)=0$, the factors of the remainder $15$ i.e. $\{-1,-3,-5,1,3,5\}$ were used as values of $x$

These were: $P(-1)$, $P(3)$ and $P(-5)$

Ok. Cool.

Now I know that this equation is the same as $(x-1)(x+3)(x-5)$. This, of course, can bring the standard form back.

Edit-1: Ok! he above is wrong. Sorry (: 

This also has factors $(x+1)(x-3)(x+5)$

I just didn't quite get why that is!

I know it works, I just don't seen to understand why the "x with opposite signal" is called factor of this polynomial. I believe my question is more about intuition than method.

Yes, I know that in order to make the equation equal to zero I need $x=1$ OR $x-3$ OR $x=-5$ above. But why do I even care about making it equal to zero? Why is it important to find the linear factors? Is it just a matter of simplification for further complexity? Is it because I can find the x-intercepts? Where is this leading me?

Anyone has any tips that might help the concept to hit me? I'm reviewing maths. I learned how to pass exams in the past, but now I am on a journey to learn real maths.

Thanks a lot

Answer 1

If $P(-1)=0$, then $P$ has some factor which is zero when $x=-1$. Which is to say $P$ has some factor which is zero when $x+1=0$. Which is to say it has $(x+1)$ as a factor.

It is this moving of the root from one side of the $=$ to the other which causes or to change into the opposite sign. Equivalently, we want the factor to be $0$ when $x$ has a specific value, which means we are forced to subtract that value from $x$ to make it $0$, which flips the sign.

Your $(x-1)(x+3)(x-5)$ doesn't bring the standard form back, as that becomes $x^3-3x-13+15$, which is not the same thing.

Answer 2

Had you used synthetic divison to evaluate your polynomial, it would have been clearer to see how evaluating a polynomial and factorization are related.

Most students are acquainted with using synthetic division to divide polynomials, but only a few realize one of its original purposes was to facilitate polynomial evaluation. (In the context of polynomial evaluation, this is referred to as Horner's rule.)

Taking $P(x)=x^3 + 3x^2 - 13x - 15$ as the example, and supposing we wish to divide this by $x-2$, the synthetic division array will look like this:

$$\begin{array}{c|rrrr}2&1&3&-13&-15\\&&2&10&-6\\\hline&1&5&-3&-21\end{array}$$

What does this say? It means $P(2)=-21$, and it also means that

$$\frac{x^3 + 3x^2 - 13x - 15}{x-2}=x^2+5x-3+\frac{-21}{x-2}$$

So, you hit two birds with one stone in synthetic division. You evaluate the polynomial at a specific value, and you also see if the polynomial has the corresponding binomial as a factor.

Now, let's see what happens if we divide by $x+1$:

$$\begin{array}{c|rrrr}-1&1&3&-13&-15\\&&-1&-2&15\\\hline&1&2&-15&0\end{array}$$

which again means that $P(-1)=0$ and

$$\frac{x^3 + 3x^2 - 13x - 15}{x+1}=x^2+2x-15+\frac{0}{x-2}=x^2+2x-15$$