Polynomial problem: Solve n^5 + 80 = 5n^4 + 16n

T

TheSheep

New member
Joined
Dec 10, 2016
Messages
5
Hello, I am a bit stuck on this problem and am having trouble solving it.

Solve: n^5 + 80 = 5n^4 + 16n, please give exact answers.

So, I started off by moving everything to the left.. n^5 - 5n^4 - 16n + 80 = 0.

Then I made all of the n's 0, and just got 80 = 0, which obviously doesn't make sense. How can I make each n = 0 and find all the possible answers for the equation?

I also noticed that an n could be factored out of the first 3 sets of numbers, so n(n^4 - 5n^3 - 16) + 80 = 0
 
TheSheep said:
So, I started off by moving everything to the left.. n^5 - 5n^4 - 16n + 80 = 0.

Then I made all of the n's 0, and just got 80 = 0, which obviously doesn't make sense.
Click to expand...

This seems like it might be a reasonable first step. However, since you got a nonsensical result, all it really tells you is that n isn't 0. You still don't know anything about what n actually is.

TheSheep said:
How can I make each n = 0 and find all the possible answers for the equation?
Click to expand...

Sorry, but I don't understand what you mean. Can you explain your thought process behind this?

TheSheep said:
I also noticed that an n could be factored out of the first 3 sets of numbers, so n(n^4 - 5n^3 - 16) + 80 = 0
Click to expand...

Factoring out a common factor is generally a good idea to start with. But, unfortunately, in this case, as you noticed, it's also not leading you anywhere, so that suggests this is maybe not the way to go after all. Are you familiar with factoring by grouping? What happens if you try to use that here? Where does that lead?
 
Hello, I am a bit stuck on this problem and am having trouble solving it.

Solve: n^5 + 80 = 5n^4 + 16n, please give exact answers.
Click to expand...
We do not give exact answers, in fact we do not give any answers. Finding answers is up to the student to find. Volunteers on the forum will assist you in getting the answers.

So, I started off by moving everything to the left.. n^5 - 5n^4 - 16n + 80 = 0.
Click to expand...

Good start

Then I made all of the n's 0, and just got 80 = 0, which obviously doesn't make sense. How can I make each n = 0 and find all the possible answers for the equation?
Click to expand...
You can't. Maybe you should use a technique that you learned in class. Maybe try factoring by grouping. In fact, try that and get back to us with your results.

I also noticed that an n could be factored out of the first 3 sets of numbers, so n(n^4 - 5n^3 - 16) + 80 = 0
Click to expand...
This is correct, but will not be helpful.
 
Last edited:
TheSheep said:
Hello, I am a bit stuck on this problem and am having trouble solving it.

Solve: n^5 + 80 = 5n^4 + 16n, please give exact answers.

So, I started off by moving everything to the left.. n^5 - 5n^4 - 16n + 80 = 0.

Then I made all of the n's 0, and just got 80 = 0, which obviously doesn't make sense. How can I make each n = 0 and find all the possible answers for the equation?

I also noticed that an n could be factored out of the first 3 sets of numbers, so n(n^4 - 5n^3 - 16) + 80 = 0
Click to expand...
Have you learned rational-root-theorem?
 
TheSheep said:
Solve: n^5 + 80 = 5n^4 + 16n, please give exact answers.

So, I started off by moving everything to the left.. n^5 - 5n^4 - 16n + 80 = 0.
Click to expand...
To learn how to solve polynomials, try here. First (as suggested earlier), apply the Rational Roots Test to get a list of potential solutions. (Then cheat with your graphing calculator to confirm which ones to try.) Then try potential zeroes in synthetic division. And so forth. ;)
 
Denis said:
You've tried n=0: no go!

Try others: n=-1, n=1, n=-2, n=2........

Seems to be what your teacher wants...
Click to expand...

So, I've taken your answer into consideration (as it was the only answer I could understand), and I cam up with the solution:

public class PolySolver {

public static void main(String[] args) {

int k = 80;
int j = 5;
int i = 16;
int m =5;
int answer;

//tests all whole numbers between -100 and 100
for ( int a = -100; a < 100; a++ ){
int c = ( a * a * a * a * a ) + k;
int b = m * ( a * a * a * a ) + i * (a);

if ( c == b )
{
answer = a;
System.out.println(answer);
}// end of if statement
}//end of for loop
}// end of main
}//end of class PolySolver




My answers were: -2, 2, 5
Thank you for all of your help, everyone.
 
Well, I mean, I guess the brute force programming solution worked in this one very specific case, but it's really not something I recommend becoming reliant on. For one thing, you most likely won't have the ability to write a program during a test, so you're out of luck on that front. But the other reason is that this approach doesn't actually find all the roots of a given polynomial. It did in this case because your instructor gave you a "nice" one that happened to only have integer roots. Even if the program were changed to account for rational numbers, there's still the possibility of irrational roots, which a brute force approach is extremely unlikely to find.

I can appreciate that math can be difficult and intimidating sometimes. But my honest advice is to really double down on your study efforts. Both Stapel and myself have provided you links to Purplemath, which I think is a good resource for learning how to factor polynomials. Khan Academy is another one - I've found that Sal Khan has a gift at explaining complicated concepts in an easy to grasp way. It may not be easy, but learning this material and really having a deep understanding of it will prove to be it's own benefit, both in terms of getting a good grade in this class, but also as a resource in future math classes which will build on this knowledge. Good luck. :)
 
TheSheep said:
Hello, I am a bit stuck on this problem and am having trouble solving it.

Solve: n^5 + 80 = 5n^4 + 16n, please give exact answers.

So, I started off by moving everything to the left.. .n^5 - 5n^4 - 16n + 80 = 0
Then I made all of the n's 0, and just got 80 = 0, which obviously doesn't make sense. How can I make each n = 0 and find all the possible answers for the equation?

I also noticed that an n could be factored out of the first 3 sets of numbers, so n(n^4 - 5n^3 - 16) + 80 = 0
Click to expand...

One "trick" to use when you have 4 terms is to group pairs together:

\(\displaystyle n^5 - 5n^4 - 16n + 80 = 0\)

\(\displaystyle (n^5 - 5n^4) - (16n - 80) = 0\)

Then factorise each pair:

\(\displaystyle n^4(n - 5) - 16(n - 5) = 0\)

If you have a common factor eg here \(\displaystyle (n - 5)\) is a common factor, you can factorise further:

\(\displaystyle (n - 5)(n^4-16)=0\)....... Note: this method won't always work, but it's worth a try.

Now, recognising the second bracket is the diff of two squares, you should be able to finish it off. See how you go.
 
Denis said:
YUK...what an unfriendly language that is!

Here's how simple it can be using Basic:

FOR n = -100 TO 100
IF n^5+80 <> 5*n^4 + 16*n THEN NEXT n
PRINT n
NEXT n

Output: -2,2,5
Click to expand...
What happened to the other two roots? Became imagination of your figments? or is it too complex solve?
 
You must log in or register to reply here.
Top