Question

hey I need some tips on finding the zeros for polynomials!

Answer 1

If you have a polynomial of the form\[ax^2+bx+c,\]where \(a,b,c\in\mathbb{C}\), then you can use\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]to find its two roots.

Answer 2

sorry, I am having more difficulty with higher degree polynomials.

for example this one
\[m^{3}+m^{2}-2=0\]

my textbook immediately factors this to
\[(m-1)(m^{2}+2m+2)\]
which is easy to solve with the quadratic formula.
However I can't see how they did that initial factorization. Does anyone have a list of things to try to factor these things?

Answer 3

@richyw have you heard of something called the Pascal's Triangle?

Answer 4

I recommend the rational root theorem, you can look into that here:
http://www.mathwords.com/r/rational_root_theorem.htm

Answer 5

http://upload.wikimedia.org/wikipedia/commons/thumb/0/0d/PascalTriangleAnimated2.gif/220px-PascalTriangleAnimated2.gif

And what @Spacelimbus said is good as well :-)

Answer 6

*Hi5's @agentx5*

Answer 7

cool, so in this case I see that the rational roots are 1, -1, 1/2, -1/2. So now I would want to use -1? first of all how do I know it's this one, second, what do I do when I determine the root I want to use? just factor out (m-root)?

Answer 8

In this case your solution for the rational root is m=1 not m=-1 , once you know the root you can do longhand division and you will obtain the expression your textbook has mentioned above.

Answer 9

\[p(1)=1^3+1^2-2=0\]
therefore you can use longhand division

\[(m^3+m^2-2):(m-1) = m^2+2m+2\]
which can also be written as
\[m^3+m^2-2= (m^2+2m+2)(m-1)\]
You always reduce the degree of the polynomial by one.

Answer 10

and to your other question, you 'don't' know which rational root you have to use. You have to try it out, usually they can easily be guessed or the constant factor at the end +k is not very big, hence the possible rational factors aren't too big.

Answer 11

ok thanks a bunch. I still don't quite get what you mean when you say they are easy to guess but at least I know the process. just worried about having to do this with more than 4 rational roots on a test!

Answer 12

In the most cases you do well enough when you just inspect the constant at the end of a polynomial.

Answer 13

Let me show you one more example to hit the point home.

\[p(x)=x^5-6x^3-6x^2-7x-6\]
In this case we are lucky, because the coefficient in front of the highest exponent is 1

So we have to try 'only' the following possible roots:
\[ \pm 1 \pm2 \pm 3 \pm 6\]
So you see, even if this method seems a bit disappointing at first, you now know that you have 'only' these 8 values to evaluate to be a possible root.

You will see that:

\[p(3)=0\]
So you can use longhand division again to reduce the degree of this polynom by one.

Answer 14

oh ok so I can just see which rational root makes the polynomial = 0 and then if it does that is what I want to use?