How is the constant term of a polynomial equation in standard form the product of the negatives of the solutions?

Question

I cannot completely grasp this concept. Additionally, what exactly ARE the negatives of solutions of a polynomial equation? Examples would be much appreciated.

Answer 1

First, some definitions:

The "solutions of a polynomial equation" in, say, the variable $x$ is the set of values of $x$ that make that equation true. For example, the solutions of $$ x^3 -5x^2 -2x +24 = 0 $$ are $\{x=-2, x=3, x=4\}$ and we know this because $$ x^3 -5x^2 -2x +24 = (x-(-2))(x-3)(x-4) $$ Second definition: The "negatives of the solutions" is the set of values each of which is the negative of one of the values in the set of solutions. In our example, the negatives of the solutions would be $$\{x=+2, x=-3, x=-4\}$$

And now you can see what happens to make the "constant term", that is, the term not involving $x$, [which is 24 in our example] equal to the product of the negatives of the solutions. Each of the solutions $-2, 3$, and $4$ appears in our product of factors as being subtracted from x. So the constant term is the product of the negatives of the solutions.

By the way, this theorem, at least in this simple form, holds only for one-variable polynomial equations.

Answer 2

Let's look at $x^2 + 5x + 6$.

It has roots $-2, -3$, because $(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0$, and a similar thing works out for $-3$.

The product $(-2)(-3) = 6$ is exactly the constant term.

Why? Because

$$ x^2 + 5x + 6 = (x+2)(x+3) = (x - (-2)) (x - (-3)) $$ which you can verify by multiplying it out.

Now on the right, it's clear that if you make $x = -2$, the first factor will be zero; if you make $x = -3$, the second factor will be zero. In both cases, the product, i.e., the original polynomial, will be zero.

When you multiply out the right hand side, you get $$ x^2 + (-2)x + (-3)x + (-2)(-3) $$ and the only term without an $x$ in it is the last one --- the product of the two roots, $-2$ and $-3$.

This happens in general: if you have

$$ p(x) = (x-a) (x-b) \cdots (x - g) $$ for instance, then $x = a, b, c, \ldots, g$ will all be roots of the polynomial $p$. When you multiply out the expression on the right, every term will contain an $x$ except the very last one, which will be $$ a\cdot b\cdot c \cdots g $$ i.e., the product of the roots. So the constant term of the polynomial is equal to the product of the roots.

This all depends on one critical thing that you didn't mention: it's essential that the coefficient of the highest-degree term is $+1$.

For instance, if we double that original polynomial, we get $$ 2x^2 + 10x + 12 $$

The roots of this are still $-2$ and $-3$, but their product is $6$, which is NOT the constant term of the polynomial. Perhaps "standard form" for you involves the leading coefficient being $1$ or something, but I wanted to be sure you weren't being mis-led.