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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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!!!!!
Can you find a factor of -30 that makes this expression equal 0?
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.
If you MUST try to get the exact real solution, I suggest you read this...
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.Quote:
Originally Posted by SMAD
-Dan
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?