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^

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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)   (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)
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