How do I determine which intervals the polynomial is entirely negative and positive on?
- Tuesday Nov 10,2009 08:40 AM
- By diddy
- In Others
49. 2x^2-4x-3
52. -1x^2+6x-10
I know how to do it if you factor it, but these don’t factor I don’t think
How would you do it algebraically? Does the quadratic equation work?
Intervals, Polynomial, Quadratic Equation
One Comment
You should know the Factor Theorem: A polynomial P(x) has a linear factor (x-r) if and only if P(r)=0. Finding (linear) factors and finding roots are equivalent. But the endpoints of your (entire) intervals on which the polynomial is positive or negative will be roots (polynomials are defined for all reals, are continuous, and only zero is neither positive nor negative). Hence the problem of finding the largest possible intervals on which the polynomial is entirely positive or entirely negative is equivalent to find the roots, which is equivalent to finding the linear factors.
Fortunately, these are quadratic equations. You can find all real roots (hence all linear factors) easily with the quadratic formula:
If ax^2+bx+c=0,
then x=(-b+Sqrt(b^2-4ac))/2a or x=(-b-Sqrt(b^2-4ac)/2a.
That gives you two, one, or zero roots (factors), depending on whether (b^2-4ac) is positive, zero, or negative. Hence you have three, two, or one interval. To check whether a particular interval is entirely negative or entirely positive, substitute any convenient number from that interval into the polynomial and look at the sign.
There are a few advanced tools that might help when the polynomial is more complex. Descartes Rule of Signs tells you that, if you substitute (x+a) {or (a-x)} for "x" and simplify to find that the coefficients all have the same sign, then the polynomial is entirely of that sign for x>a {or x<a}.
For example, substitute (0-x) for "x" in (-1x^2+6x-10) to get: (-1x^2-6x-10). Since the coefficients are all negative, we know that this polynomial is entirely negative for all x<0. [You might also try substituting (x+3) and also (3-x). Then noticing that (-1(3)-^2+6(3)-10)<0 gives a complete solution in this case.]
So some information CAN be obtained about where the roots are NOT (locating intervals which are entirely negative or positive), without actually finding any roots.
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