Question 825719: If x= sqrt(6) is a root of x^3-5x^2-6x+30=0, use synthetic division to factor the polynomial completely and list all real solutions of the equation.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website!
sqrt(6) | 1 -5 -6 30
--------- sqrt(6) 6-5sqrt(6) -30
------------------------------------
1 -5+sqrt(6) -5sqrt(6) 0
The remainder is zero (in the lower right corner) which means our (x-sqrt(6)) divided evenly into our expression (as we knew it would since it is a root). The rest of the bottom row, "1 -5+sqrt(6) -5sqrt(6)", is the quotient.
If a polynomial with rational coefficients, like our expression, has a square root for a root, then the negative of that square root will also be a root. So we will now divide our quotient by (x-(-sqrt(6))):
-sqrt(6)| 1 -5+sqrt(6) -5sqrt(6)
--------- -sqrt(6) 5sqrt(6)
------------------------------------
1 -5 0
The quotient, "1 -5", translates into x-5. This makes the last root a 5.
So the roots of our expression are: 5, sqrt(6) and -sqrt(6) (which are all real numbers).
P.S. This is a polynomial, not a rational function. Please use an appropriate category when posting problems.
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