Question

Finding all zeros in a polynomial function:
I need help with figure out how to do this. I thought I understood it but I can't seem to get down to the right answer.

Problem: Find all zeros of the function
y = x^3 - 3x^2 - 9x
I'll put up what work I did, so that if anyone knows where I went wrong or how to fix it they can show me.. Thank you in advance!

Answer 1

Sorry for the hand writing.

Answer 2

start by setting the polynomial to zero

\[0 = x^3 -3x^2 -9x\]

next take x out as a common factor

\[0 = x(x^2 -3x - 9)\]

so now you know one of the zeros since x is a factor x = 0 is a zero

you'll need to quadratic formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

to find the other zeros of the quadratic where a = 1, b = -3 and c = -9

hope this helps.

Answer 3

you mistake is that the polynomial doesn't have a constant term. so you can't use the rational root theorem.

Answer 4

Oh, thank you. I'll try and get it down to the right answer. Will you be on long enough to check it after or help me again if i get stuck?

Answer 5

yep thats fine

Answer 6

I think I went wrong again :/ Am I to come up with a fraction?

Answer 7

After that where do you go? Do you not use any coefficients when finding zeros?
@campbell_st

Answer 8

Given a zero, F, there is a factor (x-F). You DID use the coefficients. What was it that you were putting in the Quadratic Formula?

Let's work a little harder on getting the Quadratic Formula correct.
1) The 2a is the denominator for BOTH terms, not just the radical term.
2) The square root of a positive number does NOT produce "i".

Give it another go.

Answer 9

So it would just end in x = -3 +- the square root of 45/2 with 2 under the whole thing?

Answer 10

Yes, just as campbell_st has shown it, up above.

Answer 11

Okay, what do you do after you have the quadratic? do you look at both the pictures I put up? I had thought that you did a bit of both but my teacher said not to use a calculated so I don't have that to graph with to find what to use in the synthetic division.

Answer 12

so the additional 2 zeros are

\[x = \frac{3 + 3\sqrt{5}}{2}....and... x = \frac{3 - 3\sqrt{5}}{2}\]

as well as x = 0

Answer 13

How did you get to the 3 + 3 square root of 5?

Answer 14

well its simply a case of simplifying the radical

\[\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}\]

Answer 15

Ohh I see. Okay, is that all you have to do?

Answer 16

thats it... you now have the 3 zeros...

there are numerous ways of finding zeros... but the most important is

1. look and think....

straight away I saw a factor of x in every term...

removing it gave me 1 zero and a quadratic... that I knew I could manage

good luck.

Answer 17

Okay, thank you very much. Could you please check to see that every step is there and that it is correct? @campbell_st

Answer 18

you value of b = -3

so the start of the quadratic formula is -b

so you have - (-3) which becomes 3

that's the only mistake I saw

Answer 19

Thank you.. Oh and I don't mean to be a bother but could you please check to see if one other problem I did is right? @campbell_st

Answer 20

You may also want to be a little more deliberate with your notation.

\(-3^{2}\) is that \(-(3^{2})\;or\;(-3)^{2}\)? It's not always clear what is intended with that negative out in front. More parentheses solve the problem entirely.

Answer 21

you have an error in the division

you multiplied the divisor by 2x^2

which means the 1st subtraction is incorrect...


2x^3 - 3x^2
- 2x^3 - 8x^2
--------------
5x^2

Answer 22

Don't you times the x by 2x^2 to get 2x^3?

Answer 23

thats correct... but its subtraction so

2x^3 - 2x^3 = 0

and -3x^2 - (-8x^2) = 5x^2

Answer 24

look at it this way

\[2x^2(x -4) = 2x^3 - 8x^2\]

Answer 25

you forgot the minus is 8x^2...it was left as a positive by the look of things.

Answer 26

Oh my goodness -_- I didn't see that it was times by a negative 4 to get -8. Thank you ha I mess up with those a bit:/

Answer 27

its just a case of practice and being careful.

Answer 28

Is this all right now?

Answer 29

@campbell_st

Answer 30

looks good... the quadratic quotient can be factored if necessary