Method Reasoning
- Thread starter JSmith
- Start date
What methods have you been taught to find the roots (or zeroes) of polynomials?
Factor theorem and synthetic division, graphically, positive/negative interval chart, I'm mot sure what else, there have been a lot more...
OK Well I would look at your first example, and the integer coefficients would make me immediately think about the rational root theorem. And I would then quickly see that 5 is a root.
\(\displaystyle \dfrac{x^3 - 5x^2 + 2x -10}{x - 5} = x^2 + 2 \implies (x - 5)(x^2 + 2) = x^3 - 5x^2 + 2x - 10 \implies 5\ only\ real\ root.\)
And I would go on from there.
In the other, I would look at those messy coefficients and grab a graphing calculator. And then I would continue from that point.
This question is designed to get you used to the idea that you have a lot of tools in your toolbox and can use the one that seems most efficient.
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
This exercise seems to be asking for your opinion only. As worded, this exercise does not require you to describe any particular method, and you certainly do not need to solve anything.
For example, my answer would be: "I'd use a computer, for reasons of ease and accuracy."
That's my input. :cool:
D
Deleted member 4993
Guest
Notice that the expressions are inequality - so the answer would be a domain.
I would use factor by grouping.
\(\displaystyle x^3 - 5x^2 + 2x - 10 \ = \ x^2(x - 5) \ + \ ?(x - 5)\)
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
The first thing I would do is solve the associated equation. That is, for the first problem, solve \(\displaystyle x3 - 5x2 + 2x - 10 = 0\). That will have either three (not necessarily distinct) real solutions or one real and two complex solutions.
If the three real solutions are \(\displaystyle x_0< x_1< x_2\) then we can write the inequality as \(\displaystyle (x- x_0)(x- x_1)(x- x_2)> 0\). If \(\displaystyle x< x_0\) all three factors are negative so their product is negative so \(\displaystyle x< x_0\) is not a solution. If \(\displaystyle x_0< x< x_1\), then the first term, \(\displaystyle x-x_0\) is positive but the other two are still negative and f is positive.....
Of course, you should also consider the possibilities \(\displaystyle x_0= x_1< x_2\) and \(\displaystyle x_0< x_1= x_2\) as well as \(\displaystyle x_0= x_1= x_3\(\displaystyle .
And, of course, if the equation has one real root, and two complex roots, the polynomial can be written \(\displaystyle (x- x_0)(ax^2+ bx+ c)\) where \(\displaystyle ax^2+ bx+ c\) is either always positive or always negative (why). Can you see how that will affect when f is positive or when it is negative?\)\)
If the three real solutions are \(\displaystyle x_0< x_1< x_2\) then we can write the inequality as \(\displaystyle (x- x_0)(x- x_1)(x- x_2)> 0\). If \(\displaystyle x< x_0\) all three factors are negative so their product is negative so \(\displaystyle x< x_0\) is not a solution. If \(\displaystyle x_0< x< x_1\), then the first term, \(\displaystyle x-x_0\) is positive but the other two are still negative and f is positive.....
Of course, you should also consider the possibilities \(\displaystyle x_0= x_1< x_2\) and \(\displaystyle x_0< x_1= x_2\) as well as \(\displaystyle x_0= x_1= x_3\(\displaystyle .
And, of course, if the equation has one real root, and two complex roots, the polynomial can be written \(\displaystyle (x- x_0)(ax^2+ bx+ c)\) where \(\displaystyle ax^2+ bx+ c\) is either always positive or always negative (why). Can you see how that will affect when f is positive or when it is negative?\)\)