Question

Medal and fan!

What is a quartic function with only two real zeros at x = 7 and x = 13?

Answer 1

Answer choices:

Answer 2

I don't have very much of an idea about how to solve this. I've tried everything and nothing seems to work

Answer 3

it is a bit complicated process i agree
takes awhile to solve xD

Answer 4

I know. I'm so going to fail my test >.>

Answer 5

Thanks for trying though!

Answer 6

@Directrix @Hero I understand you guys might be busy but could I have some help please? I just don't understand polynomials at all.

Answer 7

well i could help i was just saying it would take some while

Answer 8

Oh, I didn't see you on here so I figured you left. I can wait. Thanks!

Answer 9

im going to mention some facts that allows you to eliminate some of the answers while will is writing his extremely length explanation

Answer 10

first, the question is asking for quartic function, this means that the highest polynomial of the function or power must be 4

bi- 2
tri-3

you can instantly see one of the answers doesnt fit

Answer 11

Okay so
we will have to run through all the options to see what function/equation contains the 2 real solutions you listed

let's start with The function
\[y = x^4- 20x^3+92x^2-20x+91\]
1st we need to identify the factors of q and p where q is the leading coefficient and p is 91

start with \[91: \pm1 , \pm7. \pm13, \pm91\]

Now the leading coefficient is 1 therefor the factors of 1 are \[\pm 1\]

Next we will find the potential zeros of this equation using p/q

therefor the potential solutions/zeros would be \[\pm1 , \pm7. \pm13, \pm91\]

Now we will plug in the 7 and 13 (expected solutions) and see if they are real solutions to this equation

x = 7
\[y = x^4- 20x^3+92x^2-20x+91\]
y = 00

when x = 13
also y = 0

Thus the zeros 7 and 13 are real solutions to the expression

Answer 12

secondly, its given that 2 roots are x =7 and x=13
this means if you were write it down as a function it would look something like
f(x) = (x-m)(x-n)(x-7)(x-13)

now, if you completely expanded this into the \(ax^4+bx^3+cx^2+dx+e\) format

you would notice that the constant d is equal to mn*7*13

which means you can look at the answers or the constants of the answers and determine which one of them are multiples of 7*13, those that aren't are not valid answers so i think

Answer 13

Solving this is not that difficult at all. All you'd have to do is grab and calculator then plug and play. Heck, you may not even need a calculator for this.

Answer 14

i can help u

Answer 15

choice 1 is deg 5 so its not quartic, thats out
choice 4 is adding a positive number, and (x+8)^4 is going to be positive, so (x+8)^4 + 71 i never 0 or below.

that leaves choice 2 and choice 3

Answer 16

pluggin 7 in
you can see that
(7-10)^4 =3^4 = 81
(13-10)^4=3^4=81
therefore
(x-10)^4-81=0, when x=7,13

Answer 17

Yup @Hero but i meant like solving for the entire zeros

Answer 18

@Will.H in this particular case, solving it that way is beyond the scope of what is required from the student. There's an unspoken rule regarding tutoring on this site which is to show the user the time-saving approach to solving these problems as many of these students also, in addition to their classroom requrements, need to also be prepared for standardized tests.

Answer 19

Thanks so much for your help, guys! I had to go earlier and just left my account idling here, but I just came back and am very happy for the information. I'll try a few more out myself and see if I can figure it out. Thanks again!