Factoring polynomials in service of calculating limits

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mathwannabe

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Hello everybody :D

Ok, I am getting obsessed with this. I'm just gonna put my problem and show you what I mean. By the way, perhaps this belongs to basic algebra, but it involves limits so I put it in here.

I have to find a limit of a function using algebraic manipulation o n l y.

\(\displaystyle \lim_{x \rightarrow 1}\dfrac{3x^4-4x^3+1}{(x-1)^2}=\)

\(\displaystyle \lim_{x \rightarrow 1}3x^2+2x+1=6\) \(\displaystyle (x\not=1)\)

What is the problem, you may ask. Well, the problem is the way I "factored" the polynomial in the numerator. I couldn't factor it. I went on like this: in order to transform the given function into its equivalent function which is defined at \(\displaystyle x=1\) the entire expression in the denominator must be canceled, so \(\displaystyle (x-1)^2\) must be a factor of the expression in the numerator. Then I did \(\displaystyle (3x^4-4x^3+1)/(x^2-2x+1)=3x^2+2x+1\)

That is just unacceptable, for me. I can factor lesser degree polynomials just fine without the use of any algorythmic approach, but this problem really got me scarred. I just need to know how to factor polynomials like this.

Could anyone show me a step by step factorization of the polynomial in the numerator?

By the way, I don't need to do this kind of problems the way I am... I am doing them just to sattisfy my insane brain :D
 
Last edited:
mathwannabe said:
Hello everybody :D

Ok, I am getting obsessed with this. I'm just gonna put my problem and show you what I mean. By the way, perhaps this belongs to basic algebra, but it involves limits so I put it in here.

I have to find a limit of a function using algebraic manipulation o n l y.

\(\displaystyle \lim_{x \rightarrow 1}\dfrac{3x^4-4x^3+1}{(x-1)^2}=\)

\(\displaystyle \lim_{x \rightarrow 1}3x^4+2x+1=6\) \(\displaystyle (x\not=1)\)

What is the problem, you may ask. Well, the problem is the way I "factored" the polynomial in the numerator. I couldn't factor it. I went on like this: in order to transform the given function into its equivalent function which is defined at \(\displaystyle x=1\) the entire expression in the denominator must be canceled, so \(\displaystyle (x-1)^2\) must be a factor of the expression in the numerator. Then I did \(\displaystyle (3x^4-4x^3+1)/(x^2-2x+1)=3x^2+2x+1\)

That is just unacceptable, for me. I can factor lesser degree polynomials just fine without the use of any algorythmic approach, but this problem really got me scarred. I just need to know how to factor polynomials like this.

Could anyone show me a step by step factorization of the polynomial in the numerator?

By the way, I don't need to do this kind of problems the way I am... I am doing them just to sattisfy my insane brain :D
Click to expand...
What you did is exactly correct, and completely acceptable in the real world. If you suspect you know a factor - or after you have found a factor - divide it out. If there is no remainder, then the factor is correct. In this case, the denominator divided evenly into the numerator, so that factor cancels out.

The methods for factoring become more "rich" [read "confusing"] when there are higher powers. Oftentimes grouping will help ..
\(\displaystyle (3x^4 - 3x^3) - (x^3 - 1) \)
\(\displaystyle \; \; \; = 3 x^3(x - 1) - (x - 1)( x^2 + x + 1)\)
\(\displaystyle \; \; \; = (3 x^3 -x^2 -x - 1)(x - 1) \)
That shows that one power of (x - 1) is a factor, which cancels one factor of (x - 1) in the denominator.
At that point I would divide the remaining cubic by (x - 1) to see if it works. "Divide and conquer!"

Even easier, if you want to test for x=1 being a root, plug in x=1 and see if the result is 0.
\(\displaystyle 3(1)^4 - 4(1^3) + (1) = 0 \)
THEN divide out (x-1), knowing the remainder will be 0. Do it again:
\(\displaystyle 3(1)^3 - (1)^2 - (1) - 1 = 0\)

In my professional life I "always" resort to numerical methods when having to find roots of a higher order polynomial [one example is finding where a randomly chosen line intersects a torus - which is 4th order]. I use a method called Newton-Raphson to find one root, then long division to reduce the order of the polynomial to 3. Repeat for the 2nd root, and divide it out. Then I am left with a quadratic and I use the quadratic formula.

Congratulations on using your "insane brain" to have fun with math!
 
mathwannabe said:
\(\displaystyle \lim_{x \rightarrow 1}\dfrac{3x^4-4x^3+1}{(x-1)^2}=\)

\(\displaystyle \lim_{x \rightarrow 1}3x^4+2x+1=6\) \(\displaystyle (x\not=1)\)

You mean \(\displaystyle 3x^2\) for the first term, correct? You reduced the degree of the
numerator by 2.

What is the problem, you may ask. Well, the problem is the way I "factored" the polynomial
in the numerator. I couldn't factor it. I went on like this: in order to transform the given
function into its equivalent function which is defined at \(\displaystyle x=1\) Were you told
that the limit would be a constant before working the problem?
It's an indeterminate form of 0/0, but the limit of those forms are not always constants.


the entire expression in the denominator must be canceled, so \(\displaystyle (x-1)^2\) must be a
factor of the expression in the numerator. Then I did \(\displaystyle (3x^4-4x^3+1)/(x^2-2x+1)=3x^2+2x+1\)
Click to expand...

...
 
DrPhil said:
What you did is exactly correct, and completely acceptable in the real world. If you suspect you know a factor - or after you have found a factor - divide it out. If there is no remainder, then the factor is correct. In this case, the denominator divided evenly into the numerator, so that factor cancels out.

The methods for factoring become more "rich" [read "confusing"] when there are higher powers. Oftentimes grouping will help ..
\(\displaystyle (3x^4 - 3x^3) - (x^3 - 1) \)
\(\displaystyle \; \; \; = 3 x^3(x - 1) - (x - 1)( x^2 + x + 1)\)
\(\displaystyle \; \; \; = (3 x^3 -x^2 -x - 1)(x - 1) \)
That shows that one power of (x - 1) is a factor, which cancels one factor of (x - 1) in the denominator.
At that point I would divide the remaining cubic by (x - 1) to see if it works. "Divide and conquer!"

Even easier, if you want to test for x=1 being a root, plug in x=1 and see if the result is 0.
\(\displaystyle 3(1)^4 - 4(1^3) + (1) = 0 \)
THEN divide out (x-1), knowing the remainder will be 0. Do it again:
\(\displaystyle 3(1)^3 - (1)^2 - (1) - 1 = 0\)

In my professional life I "always" resort to numerical methods when having to find roots of a higher order polynomial [one example is finding where a randomly chosen line intersects a torus - which is 4th order]. I use a method called Newton-Raphson to find one root, then long division to reduce the order of the polynomial to 3. Repeat for the 2nd root, and divide it out. Then I am left with a quadratic and I use the quadratic formula.

Congratulations on using your "insane brain" to have fun with math!
Click to expand...

That is Bezouts theorem, no?
 
mathwannabe said:
That is Bezouts theorem, no?
Click to expand...
Not so elegant as that, more of a brute-force approach. But in my example the algorithm has to be fast - I am doing Monte Carlo simulation of neutron scattering instruments, and one possible optical element is a mirror which is part of a toroid. I have to project the neutron trajectory to find its next intersection point and angle if incidence - and I have to do that millions of times.

I hadn't realized till I looked it up, that "divide and conquer" is a computer-science model:
https://en.wikipedia.org/wiki/Divide_and_conquer_algorithm
 
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