Question

I need help with this problem please?
Find the four real zeros (including multiplicity) of the polynomial f(x)=x^4+3x^3-12x^2-20x+48. Determine the sum of all the zeros of f (x). (Hint: The real zeros are integers that lie between -4 and 4, inclusive.)
a. 9
b. -3
c. -5
d. none of these

Answer 1

Are you familiar with rational roots theorem?

Answer 2

Yes, sort of!

Answer 3

Actually rational roots theorem may not be needed here because they give you a hint that the real zeros are integers that lie between -4 and 4, inclusive.

All you have to do is start with x = +1. Put that into x^4+3x^3-12x^2-20x+48 and see if it goes to zero. If it does then x = +1 is a root.
If not try x = -1, then x = +2 and x = -2, x = +3, x = -3, x = +4 and x = - 4.

Answer 4

so if x = +1 is a root, then x = 1 => x-1=0 (x-1) = 0

divide your polynomial by x-1, the remainder should be 0, and keep the quotient
do the same to find a 2nd root for the quotient

Answer 5

Okay!

Answer 6

I think the answer is -3. Is that correct?

Answer 7

hmm dunno.... haven't gotten the roots yet, what roots did you get?

Answer 8

-3 is one of the roots.

Answer 9

I'm still working on it... I got -3 as a root!

Answer 10

Thanks ranga!

Answer 11

Since they tell you the roots are all integers in the range -4 to +4, the quickest way to find the roots is to just plug in the 8 integers and see if f(x) = 0.

Answer 12

Alright! I'll do that!

Answer 13

But you may have to end up doing synthetic division because one of the roots has a multiplicity of 2 I think.

Answer 14

Okay!

Answer 15

... yeah.... -3 is a root... so \(\bf x^4+3x^3-12x^2-20x+48\implies (x-3)(x^3-12x+16)\)

Answer 16

Okay, thanks!

Answer 17

Oh, wait a minute. They don't ask you to find all the roots. They give you multiple choices and only one fits the bill.

Answer 18

Oh, that's right! Sorry I didn't make the answer choices clear!

Answer 19

Then you are done! You found the answer.

Answer 20

Great! Does that mean that -3 is the answer? =)

Answer 21

yes.

Answer 22

Alright! =) Both of you were so much help! Thank you!

Answer 23

you are welcome.

Answer 24

hmmm.... I read it as .... find the 4 roots, then sum them up to get an integer for the choices

Answer 25

oops. I should read questions fully before answering. :)

Sorry back to the drudgery of finding all roots! :(

Answer 26

It's okay!

Answer 27

so +4 is another... so \(\bf x^4+3x^3-12x^2-20x+48\\ \quad \\
\implies (x-3)(x+4)(x^2+4x-4)\)

and then you'd just have to do the quadratic

Answer 28

Alright, I'm working it out : )

Answer 29

BTW, if they are just interested in the sum of the roots we can quickly find the sum of the roots without actually finding the roots.
For a polynomial, the sum of the roots will be -b/a where a is the leading coefficient and b is the coefficient of the next term.

For x^4+3x^3-12x^2-20x+48
a = 1, b = 3
-b/a = -3/1 = -3.

So the sum of the roots will be -3.

But here they expect you to actually find all 4 roots and add them up.

Answer 30

Okay, thanks for showing the work =)

Answer 31

you are welcome.

Do you know synthetic division?
As soon as you found one root, -3, you can do synthetic division, get the quotient, then find the root of the quotient by substituting various x values and as soon as you find another root to do synthetic division, finding the quotient, etc.

Answer 32

Yes I know synthetic division =)

Answer 33

Then as soon as you found the first root, do synthetic division and find the quotient:
x^4+3x^3-12x^2-20x+48
x = -3 is a root. Do synthetic division:
|
-3 | 1 +3 -12 -20 +48
| -3 0 +36 -48
|________________________________________
1 0 -12 +16 0

So the quotient is: x^3 - 12x + 16

So x^4+3x^3-12x^2-20x+48 = (x + 3)(x^3 - 12x + 16)

To find another root, you can try the few integer values on the quotient.

So try x = -4 in x^3 - 12x + 16 and see if it goes to 0.

Answer 34

Alright, I get that!

Answer 35

I'm doing x = -4 in x^3 - 12x + 16 now!

Answer 36

This is a lot of work! :( sorry it's taking so long!

Answer 37

If we put x = -4 in x^3 - 12x + 16 we get
(-4)^3 - 12(-4) + 16 = -64 + 48 + 16 = -64 + 64 = 0
Therefore, x = -4 is another root.
So do synthetic division of x^3 - 12x + 16 by -4.
(note: the x^2 term is missing and so put a zero in its place)

Answer 38

|
-4 | 1 0 -12 16
|
|_____________________________

Answer 39

Okay I got (x^3+5x^2+20x+68) remainder 288

Answer 40

Using synthetic division! ^

Answer 41

way off! Looks like you may have multiplied numbers instead of adding.

Answer 42

Oh sorry : (

Answer 43

no big deal. give it a go again. drop the 1. multiply 1 and -4 and put it under 0. add. multiply by -4, etc...

Answer 44

Okay

Answer 45

Do synthetic division of x^3 - 12x + 16 by -4:
x^3 - 12x + 16 =
x^3 + 0x^2 - 12x + 16

|
-4 | 1 0 -12 16
| -4 +16 -16
|_____________________________
1 -4 4 0

Quotient is: x^2 - 4x + 4

Answer 46

sorry my internet is crashing! It keeps taking me out of this page

Answer 47

It's okay, I'll solve the rest, thanks for all the help it is really appreciated =)

Answer 48

x^2 - 4x + 4 can be factored.
x^2 - 4x + 4 = x^2 - 2x - 2x + 4
x(x - 2) - 2(x - 2)
(x - 2)(x - 2) = (x - 2)^2

So x = 2 is another root and it has a multiplicity of 2 because of (x - 2)^2. That means the root x = 2 occurs twice.

So the original f(x) = x^4+3x^3-12x^2-20x+48 =
(x - 2)^2 * (x + 3) * (x + 4)

The four roots are: -4, -3, +2, +2

Add them up and you get: - 4 + (-3) + 2 + (-2) = -4 - 3 + 2 + 2 = -7 + 4 = -3

As we expected it to be. So the answer is b.

Answer 49

no problem. And you are welcome.