These 2:
x^4 - 3x^3 + 2x^2 -7x -11
x^3 + 3x^2 -10x -1
No calculators. The method i normally use to do these by hand doesn't seem to work, but this is summer work so i might just be forgetting some steps.
What is you "normal" method? If you are trying to factor them, you should give that up. You will not be able to do it. This is not necessarily obvious.
If f(x) = Your Quartic Expression
f(-1) > 0 and f(0) < 0 -- This means one of your roots is on (-1,0)
Further
f(4) < 0 and f(5) > 0 -- This means another of your roots is on (4,5)
You must search for them. A simple bisection method is kind of fun.
There are no more Real roots, but if you drag out those two shown above, deflating f(x) to a new polynomial of lower degree, you will be left with a quadratic that can be solved easily for the two Complex Roots.
If g(x) = Your Cubic Expression
We know that you CAN solve this expression entirely. The process is painful. You probably do not want to see it, let alone do it.
By the same examination as above, we see:
g(-5) < 0
g(-4) > 0
There must be a Real Root on (-5,-4). Why?
g(-1) > 0
g(0) < 0
There must be a Real Root on (-1,0). Why?
g(2) < 0
g(3) > 0
There must be a Real Root on (2,3). Why?
In this case, you MAY wish to go hunting for all three Real roots, but this is more effort than is necessary. If you can find just one, you can reduce the problem to a mere quadratic and simply state the other two. Warning! You must find the first to wonderful precision of you expect the other two to be of much value. Deflated polynomials are VERY sensitive to bad precision.
That seems not worth the effort, maybe i just do it with a calculator and hope she gives me some credit.
My normal method(At least i think its the normal method, i only remember because i did it that way on a similar problem on an old test i dug up, my notes aren't exactly in great condition with it being summer.)
One of the problems from the test was 3x^3 + 7x^2 +5x +6
I took the factors of 6 and put them over the factors of 3. So i had a pool of factors +-6, +-2, +-1, +-1/3, etc. I pretty much guessed and checked from there. If one of those factors worked, i could use synthetic division to get the rest. -2 worked as a factor and it was pretty easy after that.
It didn't seem to work for the other problems. I could be missing something, because i really don't remember much of this and im copying what i did on a test and trying to apply it in these two problems.
Your normal method uses the rational roots theorem which basically says that if a polynomial with integer coefficients has a rational root it is the ratio of a factor of the constant term to a factor of the coefficient of the highest power of the variable.
The important word here is if, there is no guarantee that a real root of such a polynomial is rational. Both of your polynomials in this question fail to have a rational root.
RonL
The solution of cubic and quartic equations - 2These can be solved by hand using the Cardarno/ Bombelli methods see here
But this is a lot of work.
RonL