Solving cubic with a value other than one (2 Viewers)

[prime-factor]

Suppose we wish to solve
x^3 − 5x^2 − 2x+24 = 0
given that x = −2 is a solution.

We know that (x+2) will be a facor, and we can then assume that the cubic can be factored as follows:

(x+2)(x^2+ax+b)

Then by synthetic division:
.....(X+2)(x-3)(x-4)=0, x = -2, 3, 4

BUT, say the equation given was 7x^3 − 5x^2 − 2x+24 = 0, and no solutions were given. Apart from testing simple values for x e.g. (1, -1, 2, -2.....etc), and knowing that x will not be a factor of the constant since the cubic coefficient is not one, how would I even start on such a problem?. Any help would be greatly appreciated.

Thanks in Advance.

 

[lolokay]

factor has to be in the form +-a/b, where a is a factor of 24 and b is a factor of 7

have a go at proving that in the general case (being, that in a polynomial with integer coefficients, any rational factor will be in the form +-a/b where a is a factor of the constant, b is a factor of the leading term)

 

[shaon0]

Suppose we wish to solve
x^3 − 5x^2 − 2x+24 = 0
given that x = −2 is a solution.

We know that (x+2) will be a facor, and we can then assume that the cubic can be factored as follows:

(x+2)(x^2+ax+b)

Then by synthetic division:
.....(X+2)(x-3)(x-4)=0, x = -2, 3, 4

BUT, say the equation given was 7x^3 − 5x^2 − 2x+24 = 0, and no solutions were given. Apart from testing simple values for x e.g. (1, -1, 2, -2.....etc), and knowing that x will not be a factor of the constant since the cubic coefficient is not one, how would I even start on such a problem?. Any help would be greatly appreciated.

Thanks in Advance.

You could try Rational roots theorem or sum products

 

[prime-factor]

Okay,so from my equation:

7x^3 − 5x^2 − 2x+24 = 0

Possible candidates for a rational roots will be:

+/- 1, 7 , 6 , 4 , 2 , 12 , 24 , 3 , 8

And then I would have to examine the list of these possibilities as fractions?

 

[lolokay]

it can't be 7, but it can be any of those divided by 7 ^

 

[prime-factor]

Okay.

So, the factors of the constant term will be:

1, 6 , 4 , 2 , 12 , 24 , 3 , 8

and of the leading co-efficient:

1, 7

So the factors will 'have' to be these, if there are any rational roots:

+/- 1, 6 , 4 , 2 , 12 , 24 , 3 , 8
......._____________________
1, 7

So, you are saying that because 7 must be in the denominator, this rules out the possibility of 7 as a root?

 

[lolokay]

well..it's ruled out because it's not a factor of 24

 

[prime-factor]

I plugged the equation into an online solver getting one rational root, and two complex roots.

the rational root is:

-1.35729514140789

But this is not one of the fractions that I have?.

 

[lolokay]

^ that's real, but not necessarily rational
(Remember a rational number has to be able to be written as a fraction, p/q where p and q are integers)

 

[prime-factor]

Ahhh.. You are a legend!. You're right, a rational root would have to be a/b, where integers. So, I have ruled out the use of rational roots theorem. I didn't realize that answer could also be irrational.

Okay, next strategy is...sum products?. What is that?.

 

[lolokay]

probably not any easy way to solve that cubic.
did you get it from somewhere that told you to solve it, or just make it up? (looks like you just made it up actually)
so you would need to use cubic formula or something

 

[shaon0]

Ahhh.. You are a legend!. You're right, a rational root would have to be a/b, where integers. So, I have ruled out the use of rational roots theorem. I didn't realize that answer could also be irrational.

Okay, next strategy is...sum products?. What is that?.

Don't try sum product of roots. It's crap.

 

[lolokay]

lol^

yeah only do sum/product if the question gives you some information regarding those

 

[prime-factor]

This cubic is one that I made up. I'm just tired of seeing these sorts of equations and using a graphics calculator. I'm curious as to how you'd solve them. I'm in Queensland, and unfortunately a graphics calculator is used for everything. So yeah.

 

[shaon0]

This cubic is one that I made up. I'm just tired of seeing these sorts of equations and using a graphics calculator. I'm curious as to how you'd solve them. I'm in Queensland, and unfortunately a graphics calculator is used for everything. So yeah.

Are you allowed to use Graphics calculators in tests?

 

[prime-factor]

Yes. In fact, ig you turn up without one, they'll have like 3-4 spares ready to give you. I'm in grade 12 now, and you'd be spewing if you saw our curriculum. For example, binomial probability. They don't teach permutations/combinations/combinatorics at all. You use binombdf or binomcdf on the calculator and that's it. So I've been teaching myself perm's/com's etc, as well as a bunch of other maths we'll never learn. Admitadly I can't complain, I mean getting an A here, is probably like geting a C in Sydney. Also, we have no standardized tests. Every school writes there own tests.

 

[shaon0]

Yes. In fact, ig you turn up without one, they'll have like 3-4 spares ready to give you. I'm in grade 12 now, and you'd be spewing if you saw our curriculum. For example, binomial probability. They don't teach permutations/combinations/combinatorics at all. You use binombdf or binomcdf on the calculator and that's it. So I've been teaching myself perm's/com's etc, as well as a bunch of other maths we'll never learn. Admitadly I can't complain, I mean getting an A here, is probably like geting a C in Sydney. Also, we have no standardized tests. Every school writes there own tests.

lol. By the looks of things, QLD maths isn't very good.

 

[hungwell1337]

australia is bad at maths in general

russia > china > here

 

[prime-factor]

Well...you could say that. But, the basics are covered, and for those that like maths (I do, but I'm a rare find in QLD.) you can appreciate other material that is not covered.

We have three types of maths

Maths A - You can get an A very easily

Maths B - This is your mid-level difficulty mathematics (In sydney this would be about grade 7 level lol)

Maths C - Now we're talking, some good maths here.

 

[Trebla]

wtf I thought china > russia