finding zeros of a polynomial

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Hello, whitrivenbark!

This is The Factoring Problem From **** (coming to a theater near you).

Find all the zeros of this polynomial:

\(\displaystyle x^8\,+\,2x^7\,+\,3x^6\,+\,6x^5\,-\,49x^4\,-\,98^3\,+\,45x^2\,+\,90x\)
Click to expand...
First, we can factor out an \(\displaystyle x:\;\;x\cdot(x^7\,+\,2x^6\,+\,3x^5\,+\,6x^4\,-\,49x^3\,-\,98x^2\,+\,45x\,+\,90)\)


Factor 'by grouping': .\(\displaystyle x\cdot[x^6(x\,+\,2)\,+\,3x^4(x\,+\,2)\,-\,49x^2(x\,+\,2)\,+\,45(x\,+\,2)]\)

. . . and we have: .\(\displaystyle x\cdot(x\,+\,2)\cdot[x^6\,+\,3x^4\,-\,49x^2\,+\,45]\)


By inspection, we see that \(\displaystyle x\,=\,\pm1\) are zeros.

. . . Hence, we have: .\(\displaystyle x\cdot(x\,+\,2)\cdot(x^2\,-\,1)\cdot(x^4\,+\,4x^2\,-\,45)\)

And the quartic factors: .\(\displaystyle x\cdot(x\,+\,2)\cdot(x^2\,-\,1)\cdot(x^2\,-\,5)(x^2\,+\,9)\)


Set each factor equal to zero and solve for \(\displaystyle x\).

. . . \(\displaystyle x\,=\,0\)

. . . \(\displaystyle x\,+\,2\:=\:0\;\;\Rightarrow\;\;x\,=\,\)-\(\displaystyle 2\)

. . . \(\displaystyle x^2\,-\,1\:=\:0\;\;\Rightarrow\;\;x^2\,=\,1\;\;\Rightarrow\;\;x\,=\,\pm1\)

. . . \(\displaystyle x^2\,-\,5\:=\:0\;\;\Rightarrow\;\;x^2\,=\,5\;\;\Rightarrow\;\;x\,=\,\pm\sqrt{5}\)

. . . \(\displaystyle x^2\,+\,9\:=\:0\;\;\Rightarrow\;\;x^2\,=\,\)-\(\displaystyle 9\;\;\Rightarrow\;\;x\,=\,\pm3i\)

There! . . . just as pka said.
 
One has to wonder “What could possibly be a justification of such a problem?”
I took the question at face value: someone just needed to know the roots!
I also assumed that the poster did not need to know the method.
Now Soroban did a mammoth factoring job. Indeed is a factoring problem from ****.
But I did it with a CAS that costs @$40.
The power of such programs is changing mathematics education.
I, for one, think that the laptop computer will replace the calculator in mathematics education.
Therefore, I just do not understand why we still bother with such questions.
 
For those of us who don't have such programs, Microsoft Excel also does a pretty good job of it :)

(I had about six problems like this for homework last night. Killer.)
 
“For those of us who don't have such programs, Microsoft Excel also does a pretty good job of it. (I had about six problems like this for homework last night. Killer.)”
You have just made my point!

I am not a fan of Microsoft Excel. But if you can graph it, then you can see the real roots. If you enjoy this sort of thing I would encourage you to look into a computer algebra system. Here is a good one to begin with.
http://www.learnatglobal.com/html/math_catalog_6.html
 
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