How can I find the zeros to this cubic equation?

[owtu]

4x^3 + 8x^2 + 41x + 37

 

[jackarms]

Synthetic division looks like a good bet. Although you'd have to try a lot of different roots.

 

[maajdl]

Follow these lines:

http://mathworld.wolfram.com/CubicFormula.html

Alternatively, note that -1 is one the three roots.

 

[Integral]

Graph it.

 

[Mentallic]

Synthetic division looks like a good bet. Although you'd have to try a lot of different roots.

Not really, 37 is prime, so we have the possible rational roots

\pm\frac{1,37}{1,2,4}

but since all the coefficients of the cubic are positive, we can toss out the positive roots as possible solutions so all we're left with are 6 possibilities.

 

[Borek]

37 is prime

Exactly my thought, it severely limits possibilities...

http://mathworld.wolfram.com/RationalZeroTheorem.html

Doesn't help if there are irrational roots.

 

[lendav_rott]

for any polynomial there's the Horner's method, which is basically trial and error.
ax^3 + bx^2 + cx + d = 0

I prefer substituting to remove the squared member from the equation such that
y = x + \frac{b}{3a} ( Tschirnhausen's substitution I believe)and you would end up withy^3 + py + q = 0 then apply something called the Cardano's method or solution, I'm not even sure what it's called in English tbh.

x_{1,2,3} = u - \frac{p}{3u} \Longrightarrow u^3 = - \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}fromt he back of my head, this is the simplest way to tackle any cubed polynomial.

 

[Mentallic]

lendav_rott, you should always check for rational roots first because if they exist, then the problem becomes much simpler.

 

[lendav_rott]

Ok, yes, you can find the possible candidates for rational roots and test them with the Horner's method, for example, but it's such a tedious process :( Cubics are always tedious, certainly, it will become easier once you determine there are are rational roots to the problem. I guess to each their own. Mathematica or mathcad doesn't complain if I make them do something the long way , might take a few microseconds longer :(

 

[ehild]

It does not hurt to try x=1 or x=-1 before starting some complicated process. Here the root can be only negative, and -1 works !

ehild

 

[scurty]

If, once you find a rational root like -1, you can long divide (x- (root) ) into the original cubic polynomial to reduce it to a quadratic and then solve that by using the quadratic formula in order to find the other two roots.

 

[HallsofIvy]

There are, unfortunately, no rational zeros of this polynomial so you will need to use Cardano's formula.

 

[ehild]

There are, unfortunately, no rational zeros of this polynomial so you will need to use Cardano's formula.

X=-1: 4x^3 + 8x^2 + 41x + 37 =0 ---> -4+8-41+37 = 4-4 = 0

ehild

 

[D H]

There are, unfortunately, no rational zeros of this polynomial so you will need to use Cardano's formula.

Um, no. It's a trivial matter to determine that -1 is a solution: -4+8-41+37 is zero. Or, as ehild put it earlier,

It does not hurt to try x=1 or x=-1 before starting some complicated process. Here the root can be only negative, and -1 works !

 

[Ray Vickson]

Ok, yes, you can find the possible candidates for rational roots and test them with the Horner's method, for example, but it's such a tedious process :( Cubics are always tedious, certainly, it will become easier once you determine there are are rational roots to the problem. I guess to each their own. Mathematica or mathcad doesn't complain if I make them do something the long way , might take a few microseconds longer :(

Likewise, after I press 'enter', Maple returns the three solutions before my hand can reach my coffee cup. My excuse is that I have been doing this stuff for more than 50 years and have just gotten lazy in my old age. However, I am conflicted about the suggestion of an immediate jump to the use of a CAS by somebody who is just starting out in learning the material.