SOLUTION: Solve. Remember to use the Rational Zero Theorem, Descartes' Rule of Signs, reduce to a quadratic, and then solve by using factoring or the Quadratic Formula. f(x) = x3 - 7x2 +

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve. Remember to use the Rational Zero Theorem, Descartes' Rule of Signs, reduce to a quadratic, and then solve by using factoring or the Quadratic Formula. f(x) = x3 - 7x2 +      Log On


   



Question 737608: Solve. Remember to use the Rational Zero Theorem, Descartes' Rule of Signs, reduce to a quadratic, and then solve by using factoring or the Quadratic Formula.
f(x) = x3 - 7x2 + 15x - 25
Please help I have no idea where to even begin!
Thanks for your time

Found 2 solutions by richwmiller, josgarithmetic:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
factor
(x-5) *(x^2-2x+5)
since once factor is (x-5) the real solution x=5
x^2-2x+5
use quadratic formula or complete the square for the complex solutions
x = 1+2i
x = 1-2i

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
You could begin with Rule of Signs first, just to have an idea how many positive roots and how many negative roots to expect.
For the function as written, there are three sign changes, so I would expect maybe 3 or 1 positive roots.
Checking for (-x)^3-(-x)^2-x-25=-x-x^2-x-25, no sign changes. No negative roots.
My guess is that all three roots are positive; I believe I am mishandling the use of this Rule of Signs.

Rational Roots Theorem suggests possible roots to check are 1, 5, 25, or -1, -5, -25. Starting with the positives choices would be best, based on results from Descartes Rule of Signs.

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In fact, I found just what the other tutor has; that after synthetic division to check for +5 as a root, the quotient is x^2-2x+5, which has roots 1-2i and 1+2i. ( I also used solution to quadratic formula). Note that the real component of these two complex numbers is positive.

Roots: +5, 1-2i, 1+2i.