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Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet$ $f(x)+2$ is divisible by $(x+1)^3$. Write your answer in expanded form (that is, do not factor $f(x)$).

 

I've gotten (x^3(ax^2+bx+c)+2)/(x+1)^3 must be an integer, aka f(-1)=0 but don't know where to go.

 Feb 26, 2022
 #1
avatar+117872 
+2

Hi Frosty,

Welcome back to the forum. :)

 

Why have you included all that unrendered latex?

You should make it as easy as possible for people to read and interpret your questions. wink

 Feb 26, 2022
edited by Melody  Feb 26, 2022
edited by Melody  Feb 26, 2022
edited by Melody  Feb 26, 2022
 #2
avatar+117872 
+2

Find a polynomial f(x) of degree 5 such that both of these properties hold: f(x) is divisible by x^3.

and   f(x)+2 is divisible by (x+1)^3

 

Write your answer in expanded form (that is, do not factor f(x)

 

 

 

\(f(x)=x^3(ax^2+bx+c)\\ g(x)=f(x)+2=x^3(ax^2+bx+c)+2\\ g(-1)=0\quad so\\ -1(a-b+c)+2=0\\ -a+b-c+2=0\\ \boxed {c=2+b-a}\\ g(x)=x^3(ax^2+bx+2+b-a)+2\\ g(x)=ax^5+bx^4+(2+b-a)^3+2\\ \)

 

Next I divided that by  x+1 and got

\(h(x)=ax^4+(b-a)x^3+2x^2-2x+2\)

 

I know h(x) must also be diviasable by (x+1)    so  h(-1)=0

h(-1)=a(1)+(b-a)(-1)+2(1)-2(-1)+2

h(-1)=a+(-b+a)+2+2+2

h(-1)=2a-b+6 =0

b=2a+6

 

\(h(x)=ax^4+(2a+6-a)x^3+2x^2-2x+2\\ h(x)=ax^4+(a+6)x^3+2x^2-2x+2\\\)

 

Now I divided that by   (x+1)^2 = x^2+2x+1

And put the remainder = 0 hence solving for a.

 

I will leave you do the grunt work and some of the finishing up.

 

Note: This method does work but there may be a quicker way.

 Feb 26, 2022
 #3
avatar+117872 
+1

I put a great deal of time into this answer Frosty and I do want a response from you..

Melody  Feb 26, 2022
 #4
avatar
+1

so your essentially putting ?\(h(x)=0\)?

Guest Feb 26, 2022
 #5
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0

still confused, how would you find \(a\)?

Guest Feb 26, 2022
 #6
avatar+117872 
+1

Are you Frosty or are you someone else?  

Melody  Feb 26, 2022
 #7
avatar+117872 
+1

I'm not 'essentially' doing anything.

 

If a polynomial say f(x) is divisable by   (x+k)   then  f(-k) must be zero.     REMAINDER THEOREM

[  f(-k)=the remainder and the remainder in this case will be 0  ]

 

I have used this fact to help with my solution.

Melody  Feb 26, 2022
 #8
avatar+117872 
+1

How do I fing a

 

As I have already said.

1) I already found c in terms of a and b

2) I have already divided by (x+1)  using algebraic long division, to get the  b in terms of a

3) Now I need to divide that last answer by (x+1)^2 = x^2+2x+1   Using algebraic long division

 

This will give me a remiander in 'a'.  I know the remainder has to be 0 so put it equal to 0 and solve for a.

 

Bingo I will have my answer.

Melody  Feb 27, 2022

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