1. Aug 11, 2016

### Celestion

1. The problem statement, all variables and given/known data

The question was to find the roots of x3 - 3x2 + 4x - 2 = 0

2. Relevant equations

The first root is found by the factor theorem, substituting x=1 into the polynomial gives 0 therefore x=1 is one root and (x-1) is a factor.

3. The attempt at a solution

In the worked solution, then it says that the rest of the polynomial (i.e. what's left after dividing by (x-1)) is x2 - 2x + 2, "at sight", or alternatively by long division.

I know how to do the long division (and I can find the other two roots) but I can't see how to do the division in my head i.e. "at sight".

2. Aug 11, 2016

### Staff: Mentor

It's not obvious to me that one can easily get the other factor (the quadratic factor) by inspection. If you can get it by long division, that would be good enough for me -- someone who taught college math for over 20 years.

3. Aug 11, 2016

### Math_QED

On sight, I'd like to have that skill! Use either synthetic division or Horner's rule.

4. Aug 11, 2016

### symbolipoint

Only the strongest of visual-symbolic thinkers could do such division in their head. Most people are not normally that way.

5. Aug 11, 2016

### Math_QED

I think it is possible to perform this division at sight using Horner's rule, at least it worked for me and I don't consider myself a visual-symbolic thinker.

6. Aug 11, 2016

### symbolipoint

Good for you! I most likely would not be able to. I would just do synthetic division, ON PAPER; because for the example given, the process is fast.

7. Aug 11, 2016

### symbolipoint

Reread question posting #1 more carefully.
Still I cannot do the quadratic factorization for root finding in my head, but I can either complete the square or go directly to general solution of a quadratic equation.
x^2-2x+2=0
Roots are (2+- sqrt(4-4*2))/2
or
(2+- 2*sqrt(-1))/2
or
1+- sqrt(-1)
or
1+- i

Does "Horner's Method" help someone do this in his head?

8. Aug 11, 2016

### Math_QED

Only to find the quadratic can I perform the division. I would need a paper too to find the roots.

9. Aug 11, 2016

### Celestion

Thanks everyone. I'm not familiar with Horner's rule so I'll check that out. I can do basic factorisations like x2 +5x +6 in my head but that's about it.

10. Aug 11, 2016

### Celestion

The first term of x2 in x2 - 2x + 2 can be seen easily in your head. The last term of 2 also looks kind of obvious, is it generally true that the constant terms have to evenly divide? (i.e. that the -2 in the original cubic divided by the -1 in x-1 gives 2). This seems to make sense assuming there's no remainder (which there isn't because we already know x-1 is a factor). That only leaves one more term in the middle, perhaps there's some trick for seeing that?

11. Aug 11, 2016

### Celestion

In the worked solution, it doesn't mention factoring the quadratic to give 1-i and 1+i in your head, they use completing the square.

12. Aug 11, 2016

### Celestion

I looked up Horner's rule and it seems to mean rewriting the polynomial x3 - 3x2 + 4x - 2 = 0 as ((x-3)x + 4)x - 2 but I don't see how that helps to divide it by x-1?

13. Aug 11, 2016

### Math_QED

14. Aug 11, 2016

### SammyS

Staff Emeritus
I don't know if this qualifies for "at sight" or not.

x3 - 3x2 + 4x - 2
= x3 - x2 - 2x2 + 2x +2x -2
= x2(x - 1) - 2x(x - 1) +2(x - 1)

15. Aug 11, 2016

### Math_QED

16. Aug 11, 2016

### ehild

It was "at sight" in the OP. It might have been wrong translation from the original language.

17. Aug 11, 2016

### ehild

Was the problem and solution written in English?

18. Aug 12, 2016

### Celestion

Yes it's in English. The problem was in the Australian NSW higher school certificate extension 2 maths exam for 2001. The worked solution is in a book by Coroneos publications. What it says literally is

"Thus P(x) = (x-1)Q(x) where
Q(x) = x2 -2x +2 at sight, or see note."

Then it goes on to complete the square and find the two complex roots (1 +- i), and lists all three roots (including x=1) as being the answer to the question. Then it says

"Note 1. Q(x) may be obtained also by long division, as shown."

[and then the division working is shown]

I checked out the wikipedia page for Horner's method. It doen't look that much more simple than standard long division to me, perhaps if I was extremely familiar with it it could be done in my head but I'm not seeing it yet. Horner's method isn't in the syllabus for any of the NSW school maths (or any other subject) courses so I don't think the author would expect students to know it. I posted this originally because I wanted to make sure there wasn't anything that I've missed out on learning, and based on everyone's replies it doesn't seem like I have, other than Horner's method which I didn't know of.

19. Aug 12, 2016

### Celestion

There is a part in the question above that where some things are done with α, β, and γ (using the results like α+β+γ=-b/a) but none of that resembles anything that I can see how to pick the quadratic out of, there's nothing that divides the polynomial and none of the numbers for any of the combinations of roots that occur in the question correspond to the coefficients of x2 -2x +2. I guess it's possible that the author is referring to something in there, but if they are I can't see it.

20. Aug 12, 2016