SOLUTION: This question is from Algebra II, by Bob Jones University. It is in the chapter titled, "Quadratic Equations and Inequalities. In the lesson we were to "solve for x." I have the

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: This question is from Algebra II, by Bob Jones University. It is in the chapter titled, "Quadratic Equations and Inequalities. In the lesson we were to "solve for x." I have the      Log On


   

Question 162143: This question is from Algebra II, by Bob Jones University. It is in the chapter titled, "Quadratic Equations and Inequalities. In the lesson we were to "solve for x." I have the answer and I have tried many times to work the problem backwards, but I am missing something. Can you please explain this problem to me?
5x^3-39x^2+27x+7=0 x= -1/5,7,1
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
5x^3-39x^2+27x+7=0
graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C5x%5E3-39x%5E2%2B27x%2B7%29
The graph of the formula above shows possible zeroes at 1 and 7 and a third between 0 and -1.
5-39+27+7=0 so x=1
5*343-39*49+27*7+7=0 so x=7
(x-1)(x-7)=x^2-8x+7
(5x^3-39x^2+27x+7)/(x^2-8x+7)=5x+1 (long division)
5x+1=0
5x=-1
x=-1/5
.
Ed