SOLUTION: Solve the equation x^3-9x+10=0 , expressing non-integer roots in the form c+/-d(sqrt6) , where c and d are integers.

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Question 1094824: Solve the equation x^3-9x+10=0 , expressing non-integer roots in the form c+/-d(sqrt6) , where c and d are integers.
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
x^3-9x+10 = 0 .


1.  Notice that x= 2 is the root.


2.  According to the Remainder theorem, it means that the given polynomial of degree 3 is divided by the binomial (x+2) without a remainder.


3.  Divide  x^3-9x+10  by (x+2)   (long division).

    You will get a quadratic polynomial.


4.  Find the roots of this quadratic polynomial using quadratic formula.

That's all.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equation x^3-9x+10=0 , expressing non-integer roots in the form c+/-d(sqrt6) , where c and d are integers.
You can start here ====> 2 is an INTEGER root, so x = 2, and x - 2 = 0, so a factor is: x - 2. 
You then divide x%5E3+-+9x+%2B+10 by x - 2 to get a QUADRATIC trinomial.
Using this trinomial and the quadratic equation formula, you'll be able to get the other 2 roots.
Finally, present the other 2 roots in the requested form.