It is easy to verify that (x^2 - 3x - 1)(x - 1) = x^3 - 4x^2 + 2x + 1, so the quadratic polynomial is correct.
Ok, so I have a question involving a Polynomial and finding its roots.
Given
A. How many possible positive roots are there?
(Answer: 1)
B. How many possible negative roots are there?
(Answer:1)
C. What are the possible rational roots?
(Answer:
D. Using synthetic substitution, which of the possible rational roots is actually a root of the equation?
(Answer:I'm not sure on how to show synthetic division/substitution using [tex] tags, but my answer is that 1 is the actual real root of the equation.)
E. Use the quadratic formula to find the irrational roots of the resultant depressed equation.
(Answer: I'm a little fuzzy here.....the depressed equation I got was: . My question is, did I get the correct equation?
If I did then I am certain that I can do the rest, it is just the quadratic equation--one of the few things I actually understand.)
Thank you.
I only learnt one method of finding rational roots: taking the constant term, and factors there in, dividing by the coefficient of the highest degree term, and the factors there in. In this case, both numbers were 1, so this is how I got my answers.....
Since I am wrong and I know no other method, how can I find the missing root?
Hi NitroKnight,
You use Descartes' Rule of Signs to determine the number of positive and negative roots.
If used correctly, you will find that there will be 2 positive roots or no positive roots at all. There will be 1 negative root.
Who said all the roots had to be rational? You've found the only rational root which is 1.
Now use your quadratic formula with the depressed quadratic to find the 2 remaining "irrational" roots.
The Rational Roots Theorem will only help you find "rational" roots if they exist.
Ahhhhh......ok, now I see! As it happens, the last part of my question asked me to find the irrational root of my equation, thus I became confused because I had thought that the question asked first for rational roots and discounted the irrational roots till later.
So, the answer to part A would be 2? I was under the impression that i was neither positive nor negative, am I wrong about that?
That is what I meant, . I thought that was imaginary, and therefore neither positive or negative. Since the question asked me to find the rational roots, I believed that I had found them all. The last part of the question asked me find the irrational roots, so I figured that there were only two roots, one rational and one irrational, I see that I made an error--but I can't quite see where.
That is correct.I thought that was imaginary, and therefore neither positive or negative.
This conclusion is wrong.The last part of the question asked me find the irrational roots, so I figured that there were only two roots, one rational and one irrational
Solving the equation gives two real irrational roots . One of them is negative, the other positive.