Question

Factor: -2x^5 + 4x^4 -3x^3 + 2x -1

Answer 1

You'll probably have to use synthetic division, but you can start with x = 1, since if you put x=1 into: \[ -2x^5 + 4x^4 -3x^3 + 2x -1\]it'll equal zero.

Answer 2

Right, that is the one lucky break (or "easy" factor) in this problem. It is the technique after that step that causes the trouble. I do know of a way to factor this (full disclosure: it was a bonus question on an exam I took 3 semesters ago), but I always referred to the method as "black magic" math, and I was hoping someone knew of a legit way to go about working on it.

Answer 3

the numbers to use in the synthetic division are of the form \(\pm p/q\), where p is a divisor of the constant term (-1 in your problem )and q is a divisor of the leading coefficient (-2 in your problem).

the numerical coefficient of the term with the highest degree is the leading coefficient of the polynomial.

Answer 4

as to your problem, the numbers to try are \(\pm 1/1,\pm1/2\)

Answer 5

@sirm3d none of the other roots are rational numbers. Synth. division fails after finding x=1

Answer 6

then there are no more rational roots. maybe irrational roots.

Answer 7

If I understand you, you are recommending I use the rational root theorem? If I remember correctly (and I may just have to bang this out real quick to double-check), this ended up having to imaginary roots, two irrational roots, and the zero that agend0smith mentioned.

Answer 8

\[(x-1)( -2x^4+2x^3-x^2-x+1)\] now we just have to try to factor that mess

@Ajk correct, there's two irrational and two complex:
http://www.wolframalpha.com/input/?i=+-2x%5E5+%2B+4x%5E4+-3x%5E3+%2B+2x+-1+%3D+0

Answer 9

So, any ideas on how to go about this symbolically from the step above?

Answer 10

Maybe we can factor it by grouping..

Answer 11

the possible factors are (x^2 + bx + c) and (2x^2 +qx + r)
use c=1 and r = -1. if this fails, try
c=-1 and r=1

Answer 12

i found the factors (x^2 - x + 1) and (2x^2 -1)

Answer 13

i used (1-x) instead of (x-1) in the original polynomial

Answer 14

^ that is what wolframalpha found, @sirm3d :)

Answer 15

its maneable because the lead coefficient and constant are 2 and 1. imagine if they were 12 and 36. that's the real thrill. =)

Answer 16

That is pretty similar to the method I watched one of the adjuncts at my school solve it, pretty amazing actually. I doubt I would have ever had the insight to even start trying to solve it that way.

Answer 17

Here is the "black magic" method (which also would never have occurred to me, it was demonstrated by a guy who went through grad school back in the 70s). From the step after the 1 is divided out:

1. \[-2x ^{4} + 2x ^{3}-x ^{2}-x +1\]

2. Magic in (and out) another factor: \[-2x ^{4}+2x ^{3}-2x ^{2}+x ^{2}-x+1\]

3. \[-2x ^{2}(x ^{2}-x+1)(x ^{2}-x+1)\]

4.\[(x ^{2}-x+1)(-2x ^{2}+1)\]

And the rest is just solving for x using typical method.

Answer 18

Shouldn't step 3 read:\[ -2x ^{2}(x ^{2}-x+1)+(x ^{2}-x+1) \] You've got them multiplied together, not added.

Answer 19

Nifty method, though.