How do you divide polynomials in your head "on sight"?

[Celestion]

Homework Statement

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

Homework 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.

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".

 

[Mark44]

Homework Statement

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

Homework 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.

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".

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.

 

[member 587159]

Homework Statement

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

Homework 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.

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".

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

 

[symbolipoint]

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

 

[member 587159]

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

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.

 

[symbolipoint]

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.

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.

 

[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?

 

[member 587159]

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?

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

 

[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.

 

[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?

 

[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.

 

[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?

 

[member 587159]

Look at the examples on this page: https://en.m.wikipedia.org/wiki/Horner's_method

 

[SammyS]

Homework Statement

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

Homework 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.

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".

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)

 

[member 587159]

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)

Did you do that in your head?

 

[ehild]

Did you do that in your head?

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

 

[ehild]

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.

Was the problem and solution written in English?

 

[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.

 

[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.

 

[Celestion]

Check out the "method of inspection" on page 6. I just found this but I haven't got time to look at it properly until Monday. I'm not sure how much quicker it would be but it's the first I've seen of a section called "Avoiding polynomial long division" that begins on p5. This isn't an official part of the course afaik.

https://www.maths.unsw.edu.au/sites/default/files/hsc_tips_and_tricks_open_day_booklet_2015.pdf

 

[symbolipoint]

"at sight" is not clear. If this is not a well-known indication to do something purely with all steps kept in your head, then just do what you know and best understand, which would be divide your cubic P(x) by x-1, and that will be what to do. Polynomial division, or Synthetic division. The quadratic expression shown is what you expect as the quotient.

 

[Celestion]

I'll look at it properly later but now I highly suspect they're referring to the "Method of inspection" which was in a pdf that I just discovered, from a third party "tips and tricks" course for this subject.

It begins like this

"The method of inspection:
If you have a known factor, say from the remainder theorem,
you can simply inspect the quotient out of a polynomial instead of going through the
whole long division process."

and then goes on to explain it

 

[Celestion]

I'm quite happy doing the long division, though for tutoring other students I want to know all the "tricks" that may apply

 

[symbolipoint]

I'll look at it properly later but now I highly suspect they're referring to the "Method of inspection" which was in a pdf that I just discovered, from a third party "tips and tricks" course for this subject.

It begins like this

"The method of inspection:
If you have a known factor, say from the remainder theorem,
you can simply inspect the quotient out of a polynomial instead of going through the
whole long division process."

and then goes on to explain it

I have trouble understanding that relationship. The Remainder Theorem means that if a value is checked to see if it is a root, and the remainder is nonzero, then the function evaluated at that checked value is equal to the remainder found.

What Remainder Theorem means:
p(c)=d(c)Q(c)+r
If c is not a root of p, then p(c)=r.

 

[ehild]

I'm quite happy doing the long division, though for tutoring other students I want to know all the "tricks" that may apply

The method of inspection is equally good, and it might be a shorter way.
You know that x-1 is a factor of the original polynomial. So it can be written as a second order polynomial multiplied by (x-1).
x³-3x²+4x-2=(ax²+bx+c)(x-1)*
You do the multiplication on the right side, and then compare the coefficients of every power of x. They must be the same on both sides.
(ax²+bx+c)(x-1)=ax³+(b-a)x²+(c-b)x-c ---->
x³-3x²+4x-2=ax³+(b-a)x²+(c-b)x-c---->
a=1
b-a = -3
c-b = 4
c = 2
If you have a little practice, you do not need to write these equations. Looking at *, you see at once that a=1, as there is only one term with x³, and also c=2 as there is only one constant term on the right side. And then you can find b just looking at the products linear in x:
-bx+cx = 4x, which gives b=-2

 

[Celestion]

I have trouble understanding that relationship. The Remainder Theorem means that if a value is checked to see if it is a root, and the remainder is nonzero, then the function evaluated at that checked value is equal to the remainder found.

What Remainder Theorem means:
p(c)=d(c)Q(c)+r
If c is not a root of p, then p(c)=r.

Yep, that's to find the real integer factors which in this case is x-1, i.e. that p(1)=0. They are saying that assuming you already know your divisor is a factor, you can use this "method of inspection" to find the quotient. The rest of the explanation is in the PDF I linked to a couple of posts ago. I tried to paste it but PDF's don't paste well and every number appeared on a new line, with blank lines for the spaces and other symbols so that it was completely unreadable.

 

[Celestion]

The method of inspection is equally good, and it might be a shorter way.
You know that x-1 is a factor of the original polynomial. So it can be written as a second order polynomial multiplied by (x-1).
x3-3x2+4x-2=(ax2+bx+c)(x-1)*
You do the multiplication on the right side, and then compare the coefficients of every power of x. They must be the same on both sides.
(ax2+bx+c)(x-1)=ax3+(b-a)x2+(c-b)x-c ---->
x3-3x2+4x-2=ax3+(b-a)x2+(c-b)x-c---->
a=1
b-a = -3
c-b = 4
c = 2
If you have a little practice, you do not need to write these equations. Looking at *, you see at once that a=1, as there is only one term with x3, and also c=2 as there is only one constant term on the right side. And then you can find b just looking at the products linear in x:
-bx+cx = 4x, which gives b=-2

Yep, thanks very much, I reckon that's how they did it in the worked solution. One of the examples in the PDF (number 5) is almost the same question as my original post.

 

[Celestion]

I'll practice it next week and see how it goes.