Question 391606: I'm hoping someone can help me with this. Here is my problem:
x^3-2x^2-7x-4=0
Please note, the "3" and the "2" are exponents.
I have to a)list all possible rational roots
b)use synthetic division to test the possible rational roots and find an actual root
c)use the quotient from part b to find the remaining roots and solve the equation.
Ok...Here's what I have so far:
For part a) my POSSIBLE rational roots, obtained by dividing the factors of the constant term/lead coefficient are; {+-1.+-2,+-4}
For part b) I tested them via synthetic division and found that
{-1,2,4} are my ACTUAL roots.
Ok...HERE'S what I don't get: First, let me ask a silly question. If I have used all my POSSIBLE roots and found ACTUAL roots, how can any roots still be remaining? Also...WHICH quotient am I supposed to use??!!! I would be very grateful for help and clarification of part c of this problem! I need to know how to find the "remaining roots", then how to use it to solve my polynomial. Thank you to whomever responds!
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! +-1.+-2,+-4 are the possible roots.
The whole purpose of the sythetic division is to break the polynomial into smaller pieces so as to find the other roots and solve the equation. In this case if we can get one root then we will have a quadratic left and our troubles are over.
x^3-2x^2-7x-4 / x-4 = x^2+2x+1 This is the quotient.
x^2+2x+1=(x+1)(x+1)
So (x+1)(x+1)(x-4)=0
x=-1 and x=4 These are the zeros of the equation.
-1, -1, and 4 are the roots of the equation. It cannot have more than three roots because x^3 has the exponent 3.
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Ed
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