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.

FrostBringer Feb 26, 2022

#2**+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.

Melody Feb 26, 2022

#3**+1 **

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

Melody
Feb 26, 2022

#7**+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**+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