Find all the rational roots of:
1. f(x)=2x^4-5x^3+8x^2+4x+7
2. f(x)=x^3+4x^2-3x-5
3. f(x)=x^3+25
4. f(x)=x^4-x^3-7x^2+5x+10

Question

Find all the rational roots of:
1. f(x)=2x^4-5x^3+8x^2+4x+7
2. f(x)=x^3+4x^2-3x-5
3. f(x)=x^3+25
4. f(x)=x^4-x^3-7x^2+5x+10

anonymous · Answer

@IMStuck

imstuck · Answer

Gotcha!  I'm here!  You have to be familiar with the rational roots theorem.  Are you?

anonymous · Answer

Thanks! And not very

imstuck · Answer

Ok, here it is in a nutshell, so to speak.  ; )

imstuck · Answer

Look at your first one.  The 7 is in the "c" placeholder, and the first coefficient, the 2, is in the "d" position in our rational roots theorem.  All the possibilities for your factors for this are either c, d, or c/d, meaning...

anonymous · Answer

So 2, 7, 7/2???

imstuck · Answer

Don't forget the fact that 1 goes into both c and d as well.  BUT you have the possibility also of those being negative.  So all the possible roots are +1, -1, +7, -7, +1/2, -1/2, +7/2, -7/2.  Now that you have those as possilbilitites, you have to find which ones work.  Do you know synthetic division?  I may have already asked you that because this is the third problem of this nature that I am helping with!  So sorry if it's a repeat, but I need to know.

anonymous · Answer

No, I am not familiar with synthetic division

imstuck · Answer

Ok, I will show you synthetic division, with one scenario that "works" and one that doesn't.  Give me a sec to find one of your roots and then I'll show you.  You HAVE to know how to do synthetic division to find rational roots (unless you know how to divide long hand with polynomials?)

anonymous · Answer

I do know how to divide long hand with polynomials! I am just convinced as to how I would set it up?

imstuck · Answer

Let me do something here and then I'll show you both ways!  How's that?

anonymous · Answer

Ok perfect!

imstuck · Answer

Ok, the first one is a bad one to use because it doesn't have any real rational roots at all; they are all imaginary.  That's not too unusual with polynomials, but let's find one that does have real rational roots.  The answer to the first one is "none".

imstuck · Answer

Let me move on to the second one.  I'll be right back with you.  Hopefully this one is a better example!

imstuck · Answer

Ok...I'm back! Whew!  It turns out that ONLY one that has rational roots is the last one.  Here's how you go about finding it with synthetic division.  It's really extremely simple, much more than the long way!

anonymous · Answer

Ok! So the first three are all none?

imstuck · Answer

That's right!  The first three have no rational roots.  They are all either irrational or imaginary.  Just to make sure, I graphed them on my calculator and got the confirmation that way.  The third one is really simple to see because 25 is not a perfect cube.  That one was easy to figure!  Here's the set up for your synthetic division that DOeS work!

anonymous · Answer

Ok sounds good!

imstuck · Answer

First let's use the rational roots theorem and find our possible factors.  c = 10, and d = 1.  So the possibilities for c, d, and c/d are +1, -1, +2, -2, +5, -5, +10, -10.  Let's start simple with 1!  Here it is!  Ready or not...

imstuck · Answer

|dw:1403550486175:dw|This set up reflects the 1 in the little box.  That's your divisor.  The other numbers are the coefficients from your polynomial.  Include their signs, that's important, but most important is to have the x's in descending order.  If you're missing one you need to hold its place with a 0, just like you would in long division.  Got that?