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

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

I need to know only the real one. Can someone please help!!!!!

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- Nov 18th 2012, 01:41 AMmathhawkHelp!! How to factorise x^3 - 9x^2 + 24x - 30?
**Help! How to factorise x^3 - 9x^2 + 24x - 30?(Headbang)**

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

I need to know only the real one. Can someone please help!!!!! - Nov 18th 2012, 01:52 AMProve ItRe: Help!! How to factorise x^3 - 9x^2 + 24x - 30?
Can you find a factor of -30 that makes this expression equal 0?

- Nov 18th 2012, 11:44 AMBobPRe: 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. - Nov 18th 2012, 07:27 PMProve ItRe: 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...

- Nov 19th 2012, 03:10 PMSMADRe: 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.

- Nov 19th 2012, 04:28 PMtopsquarkRe: Help!! How to factorise x^3 - 9x^2 + 24x - 30?
- Nov 19th 2012, 05:41 PMSorobanRe: Help!! How to factorise x^3 - 9x^2 + 24x - 30?
Hello, mathhawk!

Quote:

Help! .How to factorise:

According to the key, 2 roots are imaginary and 1 root is real.

It is not factorable.

The only possible rational roots are: .

. . None of them are zeros of the polynomial.

We find that: .

Hence, there is a root (irrational) on the interval

Another possibility: a typo in the problem.

. . Could the constant term be -20 or -36?