# Math Insight

### Polynomial inequalities

It is worth reviewing some elementary but important points:

First, a person must remember that the only way for a product of numbers to be zero is that one or more of the individual numbers be zero. As silly as this may seem, it is indispensable.

Next, there is the collection of slogans:

• positive times positive is positive
• negative times negative is positive
• negative times positive is negative
• positive times negative is negative.

Or, more cutely: the product of two numbers of the same sign is positive, while the product of two numbers of opposite signs is negative.

Extending this just a little: for a product of real numbers to be positive, the number of negative ones must be even. If the number of negative ones is odd then the product is negative. And, of course, if there are any zeros, then the product is zero.

#### Solving inequalities

This can be very hard in greatest generality, but there are some kinds of problems that are very ‘do-able’. One important class contains problems like

Solve: $$5(x-1)(x+4)(x-2)(x+3)<0$$ That is, we are asking where a polynomial is negative (or we could ask where it's positive, too). One important point is that the polynomial is already factored: to solve this problem we need to have the polynomial factored, and if it isn't already factored this can be a lot of additional work. There are many ways to format the solution to such a problem, and we just choose one, which does have the merit of being more efficient than many.

We put the roots of the polynomial $$P(x)=5(x-1)(x+4)(x-2)(x+3)=5\,(x-1)\,(x-(-4))\,(x-2)\,(x-(-3))$$ in order: in this case, the roots are $1,-4,2,-3$, which we put in order (from left to right) $$\ldots < -4 < -3 < 1 < 2 < \ldots$$ The roots of the polynomial $P$ break the number line into the intervals $$(-\infty,-4),\; (-4,-3),\; (-3,1) ,\; (1,2) ,\; (2,+\infty)$$

On each of these intervals the polynomial is either positive all the time, or negative all the time, since if it were positive at one point and negative at another then it would have to be zero at some intermediate point!

For input $x$ to the right (larger than) all the roots, all the factors $x+4$, $x+3$, $x-1$, $x-2$ are positive, and the number $5$ in front also happens to be positive. Therefore, on the interval $(2,+\infty)$ the polynomial $P(x)$ is positive.

Next, moving across the root $2$ to the interval $(1,2)$, we see that the factor $x-2$ changes sign from positive to negative, while all the other factors $x-1$, $x+3$, and $x+4$ do not change sign. (After all, if they would have done so, then they would have had to be $0$ at some intermediate point, but they weren't, since we know where they are zero...). Of course the $5$ in front stays the same sign. Therefore, since the function was positive on $(2,+\infty)$ and just one factor changed sign in crossing over the point $2$, the function is negative on $(1,2)$.

Similarly, moving across the root $1$ to the interval $(-3,1)$, we see that the factor $x-1$ changes sign from positive to negative, while all the other factors $x-2$, $x+3$, and $x+4$ do not change sign. (After all, if they would have done so, then they would have had to be $0$ at some intermediate point). The $5$ in front stays the same sign. Therefore, since the function was negative on $(1,2)$ and just one factor changed sign in crossing over the point $1$, the function is positive on $(-3,1)$.

Similarly, moving across the root $-3$ to the interval $(-4,-3)$, we see that the factor $x+3=x-(-3)$ changes sign from positive to negative, while all the other factors $x-2$, $x-1$, and $x+4$ do not change sign. (If they would have done so, then they would have had to be $0$ at some intermediate point). The $5$ in front stays the same sign. Therefore, since the function was positive on $(-3,1)$ and just one factor changed sign in crossing over the point $-3$, the function is negative on $(-4,-3)$.

Last, moving across the root $-4$ to the interval $(-\infty,-4)$, we see that the factor $x+4=x-(-4)$ changes sign from positive to negative, while all the other factors $x-2$, $x-1$, and $x+3$ do not change sign. (If they would have done so, then they would have had to be $0$ at some intermediate point). The $5$ in front stays the same sign. Therefore, since the function was negative on $(-4,-3)$ and just one factor changed sign in crossing over the point $-4$, the function is positive on $(-\infty,-4)$.

In summary, we have $$P(x)=5(x-1)(x+4)(x-2)(x+3) > 0 \hbox{ on } (2,+\infty)$$ $$P(x)=5(x-1)(x+4)(x-2)(x+3) < 0 \hbox{ on } (1,2)$$ $$P(x)=5(x-1)(x+4)(x-2)(x+3) > 0 \hbox{ on } (-3,1)$$ $$P(x)=5(x-1)(x+4)(x-2)(x+3) < 0 \hbox{ on } (-4,-3)$$ $$P(x)=5(x-1)(x+4)(x-2)(x+3) > 0 \hbox{ on } (-\infty,-4)$$ In particular, $P(x)<0$ on the union $$(1,2) \cup (-4,-3)$$ of the intervals $(1,2)$ and $(-4,-3)$. That's it.

As another example, let's see on which intervals $$P(x)=-3(1+x^2)(x^2-4)(x^2-2x+1)$$ is positive and and on which it's negative. We have to factor it a bit more: recall that we have nice facts $$x^2-a^2=(x-a)\;(x+a)= (x-a)\;(x-(-a))$$ $$x^2-2ax+a^2=(x-a)\;(x-a)$$ so that we get $$P(x)=-3(1+x^2)(x-2)(x+2)(x-1)(x-1)$$ It is important to note that the equation $x^2+1=0$ has no real roots, since the square of any real number is non-negative. Thus, we can't factor any further than this over the real numbers. That is, the roots of $P$, in order, are $$-2< < 1 \hbox{ (twice!) }\;< 2$$ These numbers break the real line up into the intervals $$(-\infty,-2),\;(-2,1),\;(1,2),\;(2,+\infty)$$

For $x$ larger than all the roots (meaning $x>2$) all the factors $x+2$, $x-1$, $x-1$, $x-2$ are positive, while the factor of $-3$ in front is negative. Thus, on the interval $(2,+\infty)$ $P(x)$ is negative.

Next, moving across the root $2$ to the interval $(1,2)$, we see that the factor $x-2$ changes sign from positive to negative, while all the other factors $1+x^2$, $(x-1)^2$, and $x+2$ do not change sign. (After all, if they would have done so, then they would have be $0$ at some intermediate point, but they aren't). The $-3$ in front stays the same sign. Therefore, since the function was negative on $(2,+\infty)$ and just one factor changed sign in crossing over the point $2$, the function is positive on $(1,2)$.

A new feature in this example is that the root $1$ occurs twice in the factorization, so that crossing over the root $1$ from the interval $(1,2)$ to the interval $(-2,1)$ really means crossing over two roots. That is, two changes of sign means no changes of sign, in effect. And the other factors $(1+x^2)$, $x+2$, $x-2$ do not change sign, and the $-3$ does not change sign, so since $P(x)$ was positive on $(1,2)$ it is still positive on $(-2,1)$. (The rest of this example is the same as the first example).

Again, the point is that each time a root of the polynomial is crossed over, the polynomial changes sign. So if two are crossed at once (if there is a double root) then there is really no change in sign. If three roots are crossed at once, then the effect is to change sign.

Generally, if an even number of roots are crossed-over, then there is no change in sign, while if an odd number of roots are crossed-over then there is a change in sign.

#### Exercises

1. Find the intervals on which $f(x)=x(x-1)(x+1)$ is positive, and the intervals on which it is negative.
2. Find the intervals on which $f(x)=(3x-2)(x-1)(x+1)$ is positive, and the intervals on which it is negative.
3. Find the intervals on which $f(x)=(3x-2)(3-x)(x+1)$ is positive, and the intervals on which it is negative.