Solving x^3-3x^2-5x-1 without a Calculator

[Curd]

x^3-3x^2-5x-1 ??

x^3-3x^2-5x-1

apparently the book solved it by dividing it by -1 with synthetic division.

i don't like synthetic division and tried to use -x with regular division.

it did not work out.


how does one solve this without using a calculator?

please explain why each step is taken.

 

[Dick]

x^3-3x^2-5x-1

apparently the book solved it by dividing it by -1 with synthetic division.

i don't like synthetic division and tried to use -x with regular division.

it did not work out.


how does one solve this without using a calculator?

please explain why each step is taken.

If -1 is a root then (x-(-1))=(x+1) is a factor. Not -x. Try using regular division with (x+1).

 

[I like Serena]

Hi Curd!

If all coefficients are integers (like you have), you can apply the Rational root theorem:
http://en.wikipedia.org/wiki/Rational_root_theorem

This will give you a root IF there are "nice" roots.

If "a" is a nice root, you can reduce your polynomial by dividing by (x-a).

 

[Mark44]

x^3-3x^2-5x-1

apparently the book solved it by dividing it by -1 with synthetic division.

i don't like synthetic division and tried to use -x with regular division.

it did not work out.

Synthetic division is easy and convenient to use once you get the hang of it. It's much like regular polynomial division, but strips out all the stuff that isn't really needed.

-1 | 1 -3 -5 -1
   |___-1_-4_ 1_
     1 -4 -1 |0

If f(x) = x³ -3x² - 5x - 1, what we get from the above is f(-1). If f(-1) = 0, then (x - (-1)) = x + 1 is a factor of f(x).

Here are the steps after you write -1 outside the "box" and the coefficients of f(x) inside.
1. Bring the leftmost coefficient inside the box down to the bottom line. Here 1, the coefficient of x³ comes down.
2. Multiply that number by -1 and put it below the 2nd coefficient (which is -3).
3. Add the 2nd coefficient and the number below it, and write the sum in the bottom row.
4. Multiply that number by -1 and put it below the 3rd coefficient (which is -5).
5. Add the 3rd coefficient and the number below it, and write the sum in the bottom row.
6. Multiply that number by -1 and put it below the 4th coefficient (which is -1).

Since we ended up with 0 in the last entry of the bottom row, that means that f(-1) = 0, or equivalently, that x + 1 is a factor of f(x). The other numbers in the bottom row are useful, as well, because they tell us the coefficients of the other factor of f(x). In this case, we see that x³ -3x² - 5x - 1 = (x + 1)(x² - 4x -1).

 

[Curd]

the thing is i just don't like synthetic division. it feels uncertain to me. i prefer to do it the long way. I just don't get why my long way didn't work.


Ah, now i see the x+1 thing. how did they know to use that?

and how did you get the x+1 thing? how is it that when looking at the problem one should know to use x+1?

(if it's really obvious and i didn't notice i apologize. I'm feeling like crap.)

 

[Mark44]

Using the Rational Root Theorem that ILS mentioned, the only possible rational roots of the equation x³ -3x² - 5x - 1 = 0 are 1 and -1. It's entirely possible that there are roots that are irrational or even complex, but we want to check the easy ones first.

If 1 is a root of this equation, then x - 1 is a factor of the polynomial. If -1 is a root, then x - (-1) = x + 1 is a factor.

Dividing by -x should have been a nonstarter because of the -1 term in the polynomial. If there had been no constant term, then factoring out an x would have been a good idea.

 

[Curd]

for some reason i keep getting x^2-2x-3 with a remainder of 2 (what do you do with remainders anyway?).

the book gets x^2-4x-1

 

[Mark44]

http://en.wikipedia.org/wiki/Polynomial_long_division

 

[Curd]

i like how the book doesn't introduce the rational zeros theorem until the next section after this one. what's up with that?

and why are there so many names for the same theorem?

 

[Mark44]

Not knowing what book you're working worth, I can't comment on the organization.

i like how the book doesn't introduce the rational zeros theorem until the next section after this one. what's up with that?

and why are there so many names for the same theorem?

What names are you talking about? The names mentioned in this thread are Rational Root theorem and Rational Zero theorem. They are both talking about the same thing.

 

[eumyang]

i like how the book doesn't introduce the rational zeros theorem until the next section after this one. what's up with that?

After what section? Polynomial long division? Synthetic division? Both should appear before the Rational Zeros Theorem. Can you be more specific?

 

[Curd]

the section of the book I'm using.

 

[eumyang]

That wasn't helpful at all. What book are you using, and what is being covered in the section you are at currently?