finding rational zeros and rational roots.

basically it looks like this means the same exact thing, yes?

a) list all possible rational roots
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

x^3- 2x^2- 7x-4 = 0

p/q = 1,2,4/1

a) +or- 1,2,4 possible zeros
b)synthetic division yields 4 as the only one that works
c) ?????? what does this mean? If 1,2,4 are the possibilities, and I used synthetic and found that 1,2 didnt work, and 4 did, what does it mean by using the quotient to find the remaining roots? I thought I just did that in step b ?


my question is how do i do part c and also) I have this exact problem set instructions except that instead of roots it says rational zeros(instead of roots), actual zero(instead of root), and remaining zeros(instead of roots). Is the process the same? What is different?

This post was edited by fingerling on Feb 8 2014 07:11pm

part c)
after you do your synthetic division and you found 1 real root was 4, then you use the remaining values (your quotient) and find the remaining roots.

When you plugged in 4 into the synthetic division formula, you have number to the left of your zero, correct? That is your quotient
(i.e. say you have 2 3 6 0 as the numbers left over. your quotient is 2x^(2) + 3x + 6.
See how i got that?
Knowing this, i'm sure you know what to do from here

1 2 1 0 so

X^2 + 2x + 1

Do I do synthetic again? Not sure

don't know the process you are supposed to use (or only by another name) so cannot comment on that
as per the last statement, i would read as meaning the same

but some comment on how to get the roots of x^3- 2x^2- 7x-4 = 0 knowing that 4 is one
you need to solve (A*x^2+B*X+C)*(x-4)=0
perhaps the A,B and C come out of your process
just in case it helps, by a little thinking/playing i get (x-4)(x^2+2x+1)=0
which works out to be (x-4)(x+1)^2=0

edit:

i see you got it while i was typing

This post was edited by brmv on Feb 8 2014 07:43pm

that's one option.
anytime you have a quadratic equation and you get stuck, there's always the quadratic formula.
I would just reverse FOIL it

(x + 1)(x + 1)

then go from there

as the others have said:

Wave chicken bones over it (use whatever method you can) to reduce it to a quadratic, then use the quadratic formula to finish it.

So would the answer be the zero or what ?

For one of them I got x^2+2x+1

After I do quadratic I get 1 +/- zero

So what is the 1?

the three zeros/roots of x^3- 2x^2- 7x-4 = 0 are 4,-1,-1 because x^3- 2x^2- 7x-4=(x-4)(x+1)(x+1) so -1 is there twice

You didn't synthetically divide right if you only get 1 zero out of 3.

And I believe you should get 3 REAL roots. Because if you only get 1 REAL you should
get 2 COMPLEX. This would mean quadratic formula resulting in an "imaginary" root.

yes the only way a polynomial function can have complex roots is if they come in conjugate pairs (like a+i*b, and a-i*b).

A first order polynomial function will have 1 real root.
A 2nd order polynomial function will have 2 complex or 2 real roots.
A 3rd order polynomial function will either have 4 complex, 2 real + 2 complex, or 4 real roots.
A 4th order polynomial function will either have 4 complex + 1 real, 2 complex + 3 real, or 5 real roots.