You are watching: What is a constant term in a polynomial

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 $x$, **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.