Do you have to try every possible root when finding roots of a polynomial?

[zack.p]

When using synthetic division to test possible roots of a polynomial, do you have to try every possible root? Because it seems that would take forever on problems with large constant and coefficient values.

Also, do you have to get it down to a quadratic function for all problems like this before you can factor or will there be some polynomials that have 4 terms after trying all possible roots?


Thanks!

 

[Steven G]

When using synthetic division to test possible roots of a polynomial, do you have to try every possible root? Because it seems that would take forever on problems with large constant and coefficient values.

Also, do you have to get it down to a quadratic function for all problems like this before you can factor or will there be some polynomials that have 4 terms after trying all possible roots?


Thanks!

No, you do not have to try every possible root. After finding one root (call the root a), synthetic division will give you (x-a)*q(x). What is q(x) called, I forget! In any case start over again finding the possible roots for q(X). This new list of possible roots might be a shorter list then you previously had--so you do not have to test every possible zero you had from the start.

Here is another argument why you may not have to check every possible root. Suppose you have a 5th degree equation that has 10 possible rational roots. Suppose you test them one by one and find 5 roots before you tested every one of the 10 possible roots. Since a 5th degree polynomial can't have more than 5 roots it makes no sense to test the remaining possible roots.

 

[Ishuda]

When using synthetic division to test possible roots of a polynomial, do you have to try every possible root? Because it seems that would take forever on problems with large constant and coefficient values.

Also, do you have to get it down to a quadratic function for all problems like this before you can factor or will there be some polynomials that have 4 terms after trying all possible roots?


Thanks!

I assume you mean trying the rational root test. If you go about it that way [testing by synthetic division], yes you need to try all roots [unless you find one and reduce the polynomial]. For example suppose we have
f(x) = 6 x³ - 2 x² - 6 x + 2
We have 12 possibilities, one of which is x=1. However, rather than do a (synthetic) division, we could just try x=1 and quickly find out it was a root. If you started somewhere else, say x =1/3, you could multiply f(1/3) by 27 [ the cube of the denominator 3] to get
27 f(1/3) = 6 [27 * (1/27)] - 2 [27 (1/9)] - 6 [27 (1/3)] + 2 [27] = 6 - 6 - 54 + 54 = 0
which, IMO, is a bit easier than synthetic division.

As far as always getting it down to a quadratic, no you won't. A simple example of this is
g(x) = 3 x⁵ - 2 x⁴ + 3 x - 2 = (3 x - 2) (x⁴ + 1)

EDIT: As additional statements (after reading Jomo's reply), if you find x=1 is a root in the first example, do the synthetic division to get the factors (x-1) and q(x) for f(x),
q(x) = 6 x² + 4 x - 2
which might, in a different problem, be a polynomial of degree larger than 2.

To repeat something you may already know, had you tried the x = 1/3 solution, do the synthetic division as
f(x) / (3 x - 1)
and you will possibly find it easier. That is
f(x) / (denominator * x - numerator)

 

[Deleted member 4993]

When using synthetic division to test possible roots of a polynomial, do you have to try every possible root? Because it seems that would take forever on problems with large constant and coefficient values.

Also, do you have to get it down to a quadratic function for all problems like this before you can factor or will there be some polynomials that have 4 terms after trying all possible roots?


Thanks!

Checking every root is lot less stressful with a spread-sheet program like MS-Excel

 

[HallsofIvy]

I assume you mean trying the rational root test. If you go about it that way [testing by synthetic division], yes you need to try all roots [unless you find one and reduce the polynomial]. For example suppose we have
f(x) = 6 x³ - 2 x² - 6 x + 2
We have 12 possibilities, one of which is x=1. However, rather than do a (synthetic) division, we could just try x=1 and quickly find out it was a root. If you started somewhere else, say x =1/3, you could multiply f(1/3) by 27 [ the cube of the denominator 3] to get
27 f(1/3) = 6 [27 * (1/27)] - 2 [27 (1/9)] - 6 [27 (1/3)] + 2 [27] = 6 - 6 - 54 + 54 = 0
which, IMO, is a bit easier than synthetic division.

As far as always getting it down to a quadratic, no you won't. A simple example of this is
g(x) = 3 x⁵ - 2 x⁴ + 3 x - 2 = (3 x - 2) (x⁴ + 1)

This is not correct. The four roots of \(\displaystyle x^4+ 1= 0\) are x= 1, x= -1, x= i, and x= -i, That means that \(\displaystyle x^4+ 1= (x- 1)(x+ 1)(x- i)(x+ i)= (x- 1)(x+ 1)(x^2+ 1)\) with all real coefficients so \(\displaystyle 3x^5- 2x^4+ 3x- 2= (3x- 2)(x- 1)(x+ 1)(x^2+ 1)\). As long as a polynomial has all real coefficients, you can factor is to a product of first and second degree factors with real coefficients.

EDIT: As additional statements (after reading Jomo's reply), if you find x=1 is a root in the first example, do the synthetic division to get the factors (x-1) and q(x) for f(x),
q(x) = 6 x² + 4 x - 2
which might, in a different problem, be a polynomial of degree larger than 2.

To repeat something you may already know, had you tried the x = 1/3 solution, do the synthetic division as
f(x) / (3 x - 1)
and you will possibly find it easier. That is
f(x) / (denominator * x - numerator)

 

[Ishuda]

....A simple example of this is
g(x) = 3 x⁵ - 2 x⁴ + 3 x - 2 = (3 x - 2) (x⁴ + 1)...

This is not correct. The four roots of \(\displaystyle x^4+ 1= 0\) are x= 1, x= -1, x= i, and x= -i, That means that \(\displaystyle x^4+ 1= (x- 1)(x+ 1)(x- i)(x+ i)= (x- 1)(x+ 1)(x^2+ 1)\) with all real coefficients so \(\displaystyle 3x^5- 2x^4+ 3x- 2= (3x- 2)(x- 1)(x+ 1)(x^2+ 1)\). As long as a polynomial has all real coefficients, you can factor is to a product of first and second degree factors with real coefficients.

Halls,
First of all we were talking about the rational root test, see my original post 'I assume you mean trying the rational root test. ...', thus a complex number would not be considered for this particular problem.

Next, your list of the four solutions to x⁴+1=0 are incorrect, i.e.
(-1)⁴ + 1 = 2
(1)⁴ +1 =2
(i)⁴ + 1 = (-1)² + 1 = 2
and finally
(-i)⁴ + 1 = (-1)² + 1 = 2
The proper [complex] roots of x⁴+1 are \(\displaystyle \pm\dfrac{\sqrt{2}}{2}\, (1+i)\) and \(\displaystyle \pm\dfrac{\sqrt{2}}{2}\, (1-i)\)

 

[stapel]

When using synthetic division to test possible roots of a polynomial, do you have to try every possible root? Because it seems that would take forever on problems with large constant and coefficient values.

Do a quick graph on your calculator to figure out which possible roots would be best to "try" first.

Also, do you have to get it down to a quadratic function for all problems like this before you can factor or will there be some polynomials that have 4 terms after trying all possible roots?

Eh... In math class, if they're telling you to find the zeroes or factors, then they have to have given you something solvable. So, in practice, you'll pull out linear roots until you get something "nice", meaning a quadratic (where you may need to apply the Quadratic Formula), a sum or difference of squares, or some other do-able polynomials.

In real life, on the other hand.... :shock: