Question

Which of the following is a possible rational zero for the polynomial 3x3 + 5x2 - 4x + 4?
A. -3
B. 0
C. 23
D. 34

Answer 1

i say that its b

Answer 2

well did you try simplifying the polynomial??

Answer 3

Why? Can you explain what you did?

Answer 4

well you have this equation
3x^3 + 5x^2 - 4x + 4? because it is a polynomial you must separate the two.
(3x^3 + 5x^2) + (-4x - 4) and then you simplify x^2(3x + 5) -4(x - 1)
well then the zeros will be 0, -5/3, 1

Answer 5

None of the answers can be a zero of the polynomial.

Answer 6

So, @FriedRice how do I determine the zeroes after simplifying? What did you do to the simplified polynomial?

Answer 7

Why do you say that @John_ES ?

Answer 8

I say that the zeroes of the function,
\[3 x^3 + 5 x^2 - 4 x + 4\]
is none of the given answers.

Answer 9

first i separated the polynomial into 2 parts as i have shown above.
then i simplified it

Answer 10

then you have to make one of each x = 0

Answer 11

Yes I know that, you showed me that you got \[x^2(3x+5)-4(x-1)\]But how did you get the zeroes from that? @FriedRice

Answer 12

here an example from your question
x^2 = 0 so x = 0
(3x + 5) = 0 so x = 5/3
but because you have only -4 you dont have to do anything
then (x - 1) = 0 so x = 1

Answer 13

But neither of these numbers are zeroes of the function. You can plug in them, and you won't obtain 0.

Answer 14

They aren't real zeroes

Answer 15

They're rational zeroes

Answer 16

If you graphed this polynomial, you wouldn't see the graph intersect in any of these locations. However, they are still considered zeroes.

Answer 17

So @FriedRice I need to set each binomial to 0? (3x+5) (x-1) and x^2-4)?

Answer 18

hmmm then i have a another explaination.well because this is the rational roots theorem; essentially they must have a numerator that divides our constant (here 4) and a denominator which divides our leading coefficient (here 3)... note that 3 has divisors ±1,±3 and 4 has divisors ±1,±2,±4.

Answer 19

So would 3/4 work?

Answer 20

so the answer to what i suspect is option c

Answer 21

Well, this is the first time I know about such zeroes. I only know these,
http://www.sosmath.com/algebra/factor/fac10/fac10.html

Answer 22

Oh okay, I mixed up numerator and denominator there, sorry

Answer 23

Yes @John_ES I've been to that site too, but I don't understand it. I really just need someone to walk me through on of these

Answer 24

so do you understand now?

Answer 25

Somewhat. I just need to list the factors of the constant and leading coefficient and then make a fraction using the factors? @FriedRice

Answer 26

As I said before, rational zeros or zeros of the polinomy must be roots of the polinomy. Only to clarify, rational zeroes are roots of the function.

Answer 27

just listen to what john_es has to say but yes @graham69

Answer 28

Okay, thank you @FriedRice

Answer 29

no prob

Answer 30

And of course, the last answer of @FriedRice is the correct one. 2/3

Answer 31

This is a possible zero, but not the true zero of the function.

Answer 32

Alright, so if I had the polynomial \[4x^3+7x^2-5x+3\]I would take the divisors of three as ±1 and ±3, and the divisors of 4 as ±1 ±2 ±4. Then I just need to make a fraction out of them? @John_ES

Answer 33

In general, if you have a polinomy of the type,
\[ax^3+\ldots+c=0\]Then the possible zeros are divisors of c/a (positives and negatives).

Answer 34

Yes, you can do as you said.

Answer 35

Awesome. Thank you so much @John_ES

Answer 36

You're welcome.