SOLUTION: 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.

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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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