ns cannot fully grasp this concept. Additionally, what specifically ARE the negatives of options of a polynomial equation? instances would be lot appreciated.

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First, part definitions:

The "solutions the a polynomial equation" in, say, the variable $x$ is the set of worths of $x$ that make that equation true. Because that example, the remedies 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 worths each of i m sorry is the an unfavorable of one of the worths in the collection of solutions. In ours example, the negatives of the options would be $$x=+2, x=-3, x=-4$$

And currently you have the right to see what happens to make the "constant term", that is, the ax not involving $x$, same to the product the the negatives of the solutions. Each of the solutions $-2, 3$, and $4$ appears in our product of determinants as gift subtracted from x. For this reason the consistent term is the product of the negative of the solutions.

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

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reply Aug 15 "17 in ~ 22:29

note FischlerMark Fischler
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Let"s look in ~ $x^2 + 5x + 6$.

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

The product $(-2)(-3) = 6$ is exactly the consistent 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 do $x = -2$, 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 initial polynomial, will certainly be zero.

When girlfriend multiply out the ideal hand side, you get$$x^2 + (-2)x + (-3)x + (-2)(-3)$$and the just term there is no an $x$ in it is the critical one --- the product of the two roots, $-2$ and also $-3$.

This wake up 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 $p$. Once you multiply out the expression top top the right, every term will contain one $x$ other than the an extremely last one, which will be $$acdot bcdot c cdots g$$i.e., the product the the roots. So the consistent term of the polynomial is equal to the product the the roots.

This all counts on one an important thing the you didn"t mention: it"s necessary that the coefficient the the highest-degree term is $+1$.

For instance, if we twin that initial polynomial, we get$$2x^2 + 10x + 12$$

The root of this are still $-2$ and $-3$, but their product is $6$, i beg your pardon is no the consistent term of the polynomial. Perhaps "standard form" because that you requires the leading coefficient gift $1$ or something, yet I wanted to be certain you weren"t being mis-led.