# Thread: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

1. ## Help!! How to factorise x^3 - 9x^2 + 24x - 30?

Help! How to factorise x^3 - 9x^2 + 24x - 30?

It seems according to the key, 2 roots are imaginary and 1 root is real.

2. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

Can you find a factor of -30 that makes this expression equal 0?

3. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

It doesn't factorise.
A sketch graph shows that there is a single real root to the right of x = 4,
and a table of values narrows it to between x = 5 and x = 6.
You can get a better approximation if you wish, but you are then into your favourite numerical method.

4. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

If you MUST try to get the exact real solution, I suggest you read this...

5. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

Well if you have something in the form of ax^3+bx^2+cx+d=0 you can try doing synthetic division by x-1 and that will give you (x-1)q(x)+r=0.

6. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

Well if you have something in the form of ax^3+bx^2+cx+d=0 you can try doing synthetic division by x-1 and that will give you (x-1)q(x)+r=0.
And this does what? I don't see where you are headed here.

-Dan

7. ## Re: Help!! How to factorise x^3 - 9x^2 + 24x - 30?

Hello, mathhawk!

Help! .How to factorise: $\displaystyle f(x) \:=\:x^3 - 9x^2 + 24x - 30\,?$
According to the key, 2 roots are imaginary and 1 root is real.

It is not factorable.

The only possible rational roots are: .$\displaystyle \pm1,\:\pm2,\:\pm3,\:\pm5,\:\pm6,\:\pm10,\:\pm15, \:\pm30$
. . None of them are zeros of the polynomial.

We find that: .$\displaystyle \begin{Bmatrix}f(5) &=& \text{-}10 \\ f(6) &=& +6 \end{Bmatrix}$

Hence, there is a root (irrational) on the interval $\displaystyle (5,\,6).$

Another possibility: a typo in the problem.
. . Could the constant term be -20 or -36?