Question

Find the potential real zeros (roots) of the following equation.

f(x) = 4x^5 - 16x^4 + 17x^3 - 19x^2 + 13^x - 3 = 0

Answer 1

Did you try completing the square method? and can we use calculus?

Answer 2

I'm not sure what the square method is.. & I'm in pre-calc, so I'm sure you can. I'm doing my homework & my lesson didn't explain it well so I'm just trying to understand it in whatever way works

Answer 3

I don't think that method will work anyway. I need to think about pre-calc a second

Answer 4

Okay, that's fine. Take your time! Thanks for helping, by the way :)

Answer 5

Okay this is going to get messy. First, we will look at the most common evident roots. -1,0,1

Answer 6

When none of these is a root we use the rational root theorem to give us a list of other possible roots.

Answer 7

Am I teaching something new, or is this ringing a bell like it was mentioned in class

Answer 8

It rings a bell, but not like a loud church bell. Just a tiny one..

Answer 9

ok, well I plugged in, and you prob did too, and we now know that 0,1,-1 are not roots. So I'm going to explain the Rational Root Theorem

Answer 10

Yeah, I did. & Your right.. Ugh.. & Okay!

Answer 11

\[4x ^{5}-16x ^{4}+17x ^{3}-19x ^{2}+13x-3\]we create sets called say p and q. Let p be the set of numbers that are factors of 4(which is the leading coefficient in the first term of the problem) and let q be the set of factors of the last term in the problem.

Answer 12

p:{1,-1,2,-2,4,-4} q:{1,-1,3,-3}

Answer 13

are you with me so far

Answer 14

Yes, & this rings a bell. & so p is always the coefficient of the first term or the term with the highest exponent?

Answer 15

& q is the constant?

Answer 16

And this theorem says all the possible roots are p/q

Answer 17

Wait, so it's the first term?

Answer 18

Okay, I'm with you. & They you do p/q and whatever that gives you is your potential zeros & to see if they are zeros you just substitute them in?

Answer 19

Well I jut have to find the potential zeros, so wouldn't I be done?

Answer 20

wait @cutie.patootie I made a mistake

Answer 21

I switched q and p

Answer 22

forgive me, but q is the leading term's coefficient and p is the constant

Answer 23

No, that's fine. No worries!

Answer 24

so it will be constant/coefficient for the possible solutions

Answer 25

Okay, I gotcha. Thanks so much :) You were a tremendous help!

Answer 26

yup