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Is there a quick way of determining where a polynomial is positive/negative without actually plugging values?

Say you have a polynomial

$$1) f(x)=(x+a)(x+b)$$

or

$$2) f(x)=(x-a)(-x+b)(x-c)$$

How can one quickly know whether function $f(x)$ is positive or negative without plugging in random interval values between the roots?

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2 Answers 2

Looking at $(x+a)(x+b)$, if $x$ is "large" then the polynomial is positive. Now imagine $x$ decreasing. Provided $a\neq b$, once $x$ crosses the larger of $-a, -b$, $f(x)$ will change sign. Once $x$ crosses the other one, it will change sign again.

This strategy needs to be modified if any of the roots are repeated. For example, $f(x)=(x-2)^5(x-3)^4$. This is positive for "large" $x$. Crossing below $3$ the sign does not change, because $4$ is even, so $f(x)$ is positive in $(2,3)$. Crossing below $2$ the sign does change, because $5$ is odd, so $f(x)$ is negative on $(-\infty, 2)$.

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Sure. If you have those factorizations, then it's not so bad. You want to look at the sign of each individual term.

In the case of $(1)$, we know that $f(x)$ will be positive when $(x+a)$ and $(x+b)$ are the same sign, and negative when they are different. When is $x + a < 0$? Well, when $x < -a$. When is $x + b < 0$? Similarly, when $x < -b$. Without loss of generality, assume $a > b$.

Then when $x < -a$, we also have $x < -b$. So we would have $f(x) = (x+a)(x+b) = (-)(-) = (+)$.

How about when $-a < x < -b$? Then we have $f(x) = (-)(+) = (-)$.

And when $-a < -b < x$? Then we have $f(x) = (+)(+) = (+)$.

Of course, when you have either $x = -a$ or $x = -b$, then $f(x) = 0$. And if $a = b$, $f(x) = (x+a)^{2}$, so it's always nonnegative, and zero only when $x = -a$.

Using this kind of analysis, can you try your second example?

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