finding zeros of a polynomial

[whitrivenbark]

need help to find all the zeros of this polynomial:

x^8+2x^7+3x^6+6x^5-49x^4-98^3+45x^2+90x

 

[pka]

 

[soroban]

Hello, whitrivenbark!

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

Find all the zeros of this polynomial:

$\displaystyle x^8\,+\,2x^7\,+\,3x^6\,+\,6x^5\,-\,49x^4\,-\,98^3\,+\,45x^2\,+\,90x$

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.

 

[pka]

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.

 

[Guest]

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.)

 

[pka]

“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