Cubic Equations

Sully

If your asked to find the zeros, and you are given one of them, for a cubic equation - whats the correct method to break down the equation into a quadratic? I can do most, but I came across a tricky enough question;

-x^3-2x^2+x+2

Cant seem to find another root which will give me zero, so I assume my method of breaking it into a quadratic is wrong (take out -x, divide across, move the number two outside and change its sign leaving two functions).

Vegeta

When given no roots Isn't one of the methods in the leaving cert trial and error using division?

If you are given one root you divide it into the cubed function leaving a quadratic.

Is that what you're talking about? It seems so. What root were you given and we'll have a crack

Sully

Jesus.. two long division sums in the one question? (Oblique Assymptote in part A).

The root is 1 (this topic is on part B of the question - zeros/max & min/inflection pts)

Thomas_S_Hunterson

http://www06.wolframalpha.com/input/?i=solve+-x^3-2x^2%2Bx%2B2 :P

/edit: but yea as vegeta said, given 1 is a root, (x-1) is a factor so divide and get a quadratic.

Sully

Sean_K wrote: »

http://www06.wolframalpha.com/input/?i=solve+-x^3-2x^2%2Bx%2B2 :P

Shame it doesn't show the steps taken!

Vegeta

as 1 is a root, divide your original equation by (x-1)

This will give -x^2-3x-2

Put it equal zero, change sign and it simplifies to (x+2)(x+1),

So roots are 1, -1, -2

Sully

Vegeta wrote: »

as 1 is a root, divide your original equation by (x-1)

This will give -x^2-3x-2

Put it equal zero, change sign and it simplifies to (x+2)(x+1),

So roots are 1, -1, -2

x-1/-x^3-2x^2+x+2

Nasty enough.

Vegeta

Sully wrote: »

x-1/-x^3-2x^2+x+2

Nasty enough.

Yeah its a nasty little fecker.

They're worse if not given a root as you have to just start trying different roots (dividing) until you get one that works. That could mean 2 or 3 division attempts. The division itself is easy but with your example I found myself having to be extremely careful not to mess a sign up.

Thomas_S_Hunterson

/complicates things a bit:
http://mathworld.wolfram.com/CubicFormula.html

You can always use the factor theorem to find one root if you're starting off with no info about the cubic.

Sully

Vegeta wrote: »

as 1 is a root, divide your original equation by (x-1)

This will give -x^2-3x-2

Put it equal zero, change sign and it simplifies to (x+2)(x+1),

So roots are 1, -1, -2

When I worked it out I got;

-x^2+3x+4

See. Maths and me are worst enemies. You would think after studying this for years, id have a better grasp eh!

Sully

Sully wrote: »

When I worked it out I got;

-x^2+3x+4

See. Maths and me are worst enemies. You would think after studying this for years, id have a better grasp eh!

[latex]\displaystyle\mbox{-x^2 - 3x - 2}[/latex] is the correct answer for division by (x-1), anyway.

A simple way to check your answers is to multiply what you divided by, by your result. It'll quickly let you know if you've gone wrong somewhere.

Fremen

Here's a quick trick. Won't always work, but it'll save you lots of time when it does.

In this case, you know

(-1)(x -1)(x - a)(x - b) = (whatever your polynomial is)

You can multiply out the brackets, and equate the constant terms. In this case,

(-1)(-1)(-a)(-b) = 2

or
ab = 2

Now, guess at choices for a and b. In an exam-type situation, the coefficients will be chosen to give nice solutions, so this will often work.

The obvious choices are (a=1 b=2) , (a=Sqrt(2), b = Sqrt(2)), (a=-1, b=-2) and (a= -Sqrt(2), b = -Sqrt(2)). It'd take one minute to check these, and would save you time and hassle.

Sully

-JammyDodger- wrote: »

[latex]\displaystyle\mbox{-x^2 - 3x - 2}[/latex] is the correct answer for division by (x-1), anyway.

A simple way to check your answers is to multiply what you divided by, by your result. It'll quickly let you know if you've gone wrong somewhere.

Strange..

Whats the easiest way of writing out my solution?

Sully

Sully wrote: »

Strange..

Whats the easiest way of writing out my solution?

Of writing your solution?

Are you referring to the LaTeX code I used?

Sully

Nevermind, got it right - daft use of signs made a complete balls of it.

Cheers

Just a tip for you (which makes things easier):

The equation which you gave was [latex]\displaystyle\mbox{- x^3 - 2x^2 + x + 2 = 0}[/latex], you can multiply all of that by -1, giving [latex]\displaystyle\mbox{x^3 + 2x^2 - x - 2 = 0}[/latex]. This seems like a pretty small change, but it makes the division a lot easier if the [latex]\displaystyle\mbox{x^3}[/latex] and [latex]\displaystyle\mbox{x^2}[/latex] terms are positive (at least one of them, anyway).

Sully

-JammyDodger- wrote: »

Just a tip for you (which makes things easier):

The equation which you gave was [latex]\displaystyle\mbox{- x^3 - 2x^2 + x + 2 = 0}[/latex], you can multiply all of that by -1, giving [latex]\displaystyle\mbox{x^3 + 2x^2 - x - 2 = 0}[/latex]. This seems like a pretty small change, but it makes the division a lot easier if the [latex]\displaystyle\mbox{x^3}[/latex] and [latex]\displaystyle\mbox{x^2}[/latex] terms are positive (at least one of them, anyway).

Thanks. I'm not the greatest at maths and trying (yet again) to complete a college Applied Calculus exam. I usually know how to do questions, but often get stumped if its a hard enough sum to work out. If I dont pass this, I wont get into fourth year and iv actually already repeated secound year to pass this exam with no luck.

Sully

Sully wrote: »

Thanks. I'm not the greatest at maths and trying (yet again) to complete a college Applied Calculus exam. I usually know how to do questions, but often get stumped if its a hard enough sum to work out. If I dont pass this, I wont get into fourth year and iv actually already repeated secound year to pass this exam with no luck.

Well applied calculus wouldn't be the easiest, anyway. I'm sure you'll pass it, just remember that if you're stuck at any question this forum is a great resource (not only for solutions, but there are many people here who would, if you wanted, explain the questions in great detail). Best of luck with it in the end.

Sully

Sully wrote: »

If your asked to find the zeros, and you are given one of them, for a cubic equation - whats the correct method to break down the equation into a quadratic? I can do most, but I came across a tricky enough question;

-x^3-2x^2+x+2

Cant seem to find another root which will give me zero, so I assume my method of breaking it into a quadratic is wrong (take out -x, divide across, move the number two outside and change its sign leaving two functions).

Just looking at the above again. In the above example a root was given but in others similar I must determine the solution. I assume I don't simplify into a qudartic and get roots that way and instead must try different numbers through trial and error?

With regards to simplification, its not possible if there is a whole number in the equation as in this question, right?

Exams tmrow anyway, so wish me luck :P