1. ## Polynomial inequalities

I am currently reading about polynomial inequalities and came across this statement that I do not particularly understand: "To solve a polynomial inequality, you can use the fact that a polynomial can change signs only at its zeros."

What do they mean by this?

2. ## Re: Polynomial inequalities

For example:
f(x)=x^2-4x+3
You want to solve: x^2-4x+3<0
You say: x^2-4x+3=0 <-> (x-1)(x-3)=0
You have solved it for f(x)=0 but you want to know the numbers where f(x)<0 so you draw a picture of the x-axis and mark the point where f(x) becomes zero so:
x---------1-------3--------
f(x) -----0-------0--------
You can now determine where f(x)<0 or f(x)>0 so a polynomial can change sings only at its zeros.

3. ## Re: Polynomial inequalities

Another question: are the critical numbers the numbers that make the polynomial zero?

4. ## Re: Polynomial inequalities

yes. they define the intervals of the inequality. substitute a test point from each interval into the inequality and solve. if the result inequality is true then that interval is part of the solution set. if it is false then it is not part of the solution set.

5. ## Re: Polynomial inequalities

Now I am reading about rational inequalities and it says: "...use the fact that the value of a rational expression can change sign only at its zeros and its undefined values." Could someone possibly explain this in a more simpler way?

6. ## Re: Polynomial inequalities

Originally Posted by Bashyboy
Now I am reading about rational inequalities and it says: "...use the fact that the value of a rational expression can change sign only at its zeros and its undefined values." Could someone possibly explain this in a more simpler way?
Suppose we have

$\displaystyle f(x)=\frac{x-2}{x-3}$

This changes sign at x=2, which is the zero.
Another sign change occurs on both sides of x=3,
the value of x not part of the function's domain.
If x is just to the left of 3 and x>2, the denominator is negative
and the graph approaches negative infinity.
If x is just to the right of 3, the denominator is positiive
and the graph goes to positive infinity.

There is a discontinuity in the graph at x=3
and the graph goes to opposite ends of the y-axis either side of x=3.

It's a sign change, but not a continuous one as we have at the zero.

7. ## Re: Polynomial inequalities

All polynomials can be factored so all rational functions can be written as a quotient of a product of linear factors:
$\displaystyle \frac{x- a_1)(x- a_2)\cdot\cdot\cdot(x- a_n)}{(x- b_1)(x- b_2)\cdot\cdot\cdot(x- b_m)}$

It is also true that the product of two negative numbers is positive and the product of two positive number is positive while the product of two numbers of different sign is negative. That means that we can determine whether or not a product or quotient is positive or negative just by counting the number of negative terms- if it is odd, the product is negative, if it is even, the product is positive. Now suppose that in such a product the term $\displaystyle x- a_i$ occurs only once and x is very close to $\displaystyle a_i$. For $\displaystyle x< a_i$, $\displaystyle x- a_i< 0$ while for $\displaystyle x> a_i$, $\displaystyle x- a_i> 0$. No other factor changes sign as we go from $\displaystyle x< a_i$ to $\displaystyle x> a_i$ so only that term changes sign. The number of negative terms has changed by one and so has gone from even to odd or vice versa. In either case the sign of the entire product/quotient has changed.

Of course, if the factor $\displaystyle (x- a_i)$ itself occurs an even number of times, say as $\displaystyle (x- a_i)^4$, the entire function does not change sign but the point is that the only place such a product/quotient can change sign is where some individual factor does- where x changes from $\displaystyle x< a_i$ to $\displaystyle x> a_i$ which is, of course, at $\displaystyle x= a_i$.