Factorising cubic HELP!

So I have a cubic function to be f(x) = 2x^3 - x^2 - 13x - 6

I am meant to factorise this cubic function out completely so I can work out its roots. The problem is I'm not given one of the roots to start off with so that I can use long division to factor it out.

Is there a method I can use to find one of the first roots quickly without having the need to guess and predict which root could fall into the function?

I find that trying to predict which root it can be is sometimes long and more likely to cause a mistake.

Any help would be great!

Thanks!

Factor theorem

Try x = p/q
where p is the constant term, q is the coefficient of x^3
e.g x = 1,-1,2,-2,3,-3,6,-6,3/2 etc

Okay so I know the constant term to be -6 and the coefficient of x^3 is 2.
Therefore -6/2 = -3.

So I can say that (x-3) is factor of this cubic function. Am I right to say this?

yes

Thanks. I was wondering is this method always a guaranteed way of quickly working out one of the roots.
If so could this also work the same way if I was to factorise out a quartic function too?

Okay so I know the constant term to be -6 and the coefficient of x^3 is 2.
Therefore -6/2 = -3.

So I can say that (x-3) is factor of this cubic function. Am I right to say this?

This only works if the function has a rational root. For irrational/complex roots it won't work

Ahh I see. So let's suppose I had to factorise out a function raised to the fifth power. If I were to factorise this out completely using the factor theorem. I would take possible factors of the constant term and then divide by any possible factors of the coefficient of x^5. Right?

Yes

There's no easy way tbh. The best you could do is try differentiating it and see where the stationary points are to predict some roots if you try sketching a graph. Then use the intermediate value theorem after looking at the remainders when you plug in some numbers. Btw your above statement is false, it's not guaranteed to be a root but could be. And yes if you are dealing with A-level stuff, then usually the factor theorem gives the thing that you need.

So I have a cubic function to be f(x) = 2x^3 - x^2 - 13x - 6

I am meant to factorise this cubic function out completely so I can work out its roots. The problem is I'm not given one of the roots to start off with so that I can use long division to factor it out.

Is there a method I can use to find one of the first roots quickly without having the need to guess and predict which root could fall into the function?

I find that trying to predict which root it can be is sometimes long and more likely to cause a mistake.

Any help would be great!

Thanks!

If you have covered roots of polynomials then you should be familiar with the fact that if a cubic equation has roots $\alpha, \beta, \gamma$ then it can be factorised as

$(x-\alpha)(x-\beta)(x-\gamma) = 0$

$\implies x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha)x - \alpha \beta \gamma = 0$.

In your case, your cubic of $2x^3 - x^2 - 13x - 6 =0$ can be divided through by the leading coefficient to obtain

$x^3 - \dfrac{1}{2}x^2 - \dfrac{13}{2}x - 3 = 0$.


From comparing constant terms between our two cubic forms, we get that $\alpha \beta \gamma = 3$.

So if your cubic has an integer root, then it must be $\alpha \in \{ \pm 1, \pm 3 \}$. I.e. factors (and their -ve versions) of $3$.

To be certain which one of these it is, just evaluate $f(\alpha)$ for all possibilities and see for which your function is zero.

Then it's just polynomial division from there to finish off the factorisation.

I think I'm following but I don't know what you exactly mean when you mentioned 'To be certain which one of these it is, just evaluate for all possibilities and see for which your function is zero.'

Do you find the alpha value from simply picking it out all the possible factors of 3 and then plugging it into the function and then see which is equal to zero?

I'm saying you should substitute $1, -1, 3, -3$ and see for which one of these you get your $f(x) = 0$.

I simply denoted that root to be $x = \alpha$, hence why I said $f(\alpha) = 0$.

What approach would you take if there wasn't a constant term at the end of the function? Like if the the constant term was just zero. Would the rational root theorem still hold true?

What approach would you take if there wasn't a constant term at the end of the function? Like if the the constant term was just zero. Would the rational root theorem still hold true?

Then clearly one of your roots must be zero. It's a direct implication.

$\alpha \beta \gamma = 0$ so at least one of these roots must be zero.

It's a slightly easier case. You can notice it anyway without this knowledge since you can factor out an $x$ when there is no constant term.