SOLUTION: If I have the equation f(x)=3x^3-43x^2+161x-49 are the real zeros 7 and 1/3. If so how do you use the real zeros to factor the equation? And if I have the equation f(x)=2x^3-4x^

Algebra ->  Functions -> SOLUTION: If I have the equation f(x)=3x^3-43x^2+161x-49 are the real zeros 7 and 1/3. If so how do you use the real zeros to factor the equation? And if I have the equation f(x)=2x^3-4x^      Log On


   

Question 854420: If I have the equation f(x)=3x^3-43x^2+161x-49 are the real zeros 7 and 1/3. If so how do you use the real zeros to factor the equation?
And if I have the equation f(x)=2x^3-4x^2-34x+68 are the real zeros positive and negative sqrt(17) and 2? And how do you use the real zeros to factor f?
HELP!!!!!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

f(x)=3x^3-43x^2+161x-49    |Yes x = 7 is a root

Using synthetic Division to find the other two

7	3	-43	161	-49

		21	-154	49

	3	-22	7	 0

    3x^2 - 22 + 7

   ( 3x - 1)(x - 7) = 0 , x = 7 , x = 1/3

f(x)=3x^3-43x^2+161x-49 = (x-7)(x-7)(x - 1/3)



f(x)=2x^3-4x^2-34x+68  x = 2 is a root

Using synthetic Division to find the other two

2	2	-4	-34	68

		4	0	-68

	2	0	-34	0

        2x^2 = 34 = 0

         x = ± √17

f(x)=2x^3-4x^2-34x+68 = (x-2)(x + √17)( x -√17)