Please Help ASAP!!! Polynomial Division

ok so for the first part of your question  f(x) is divisible by $x^3$, the answer is

*THIS IS ALL HEUREKA'S WORK ABOVE*

Find a polynomial \(f(x)\) of degree 5 such that both of these properties hold:

\(f(x)\) is divisible by \(x^3\).
\(f(x)+2\)  is divisible by \((x+1)^3\).

1. \(\bf{\text{ $\mathbf{f(x)}$ is divisible by $\mathbf{x^3}~$ ?}}\)

\(\begin{array}{lcll} \text{Let $f(x) = a(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$ is a polynomial of degree $5$.} \\ \text{If f(x) is divisible by $x^3$, than $x_1=x_2=x_3=0 $} \\ \qquad f(x) = a(x-0)(x-0)(x-0)(x-x_4)(x-x_5) \\ \qquad \boxed{f(x) =ax^3(x-x_4)(x-x_5)} \text{ This polynom is divisible by $x^3$} \\ \text{expand to:} \\ \qquad f(x)= ax^5 - a(x_4+x_5)x^4 + ax_4x_5x^3 \qquad \text{Let $A=a$, $~B=a(x_4+x_5)$, and $~C = ax_4x_5 $ } \\ \text{finally we have:} \\ \qquad \boxed{f(x) = Ax^5+Bx^4+Cx^3 \qquad (1) } \text{ This polynom is divisible by $x^3$} \\ \end{array}\)

2. \(\bf{\text{ $\mathbf{f(x)+2}$ is divisible by $\mathbf{(x+1)^3}~$ ?}}\)

\(\begin{array}{lcll} \text{Say $P(x) $ is a polynom divisible by $(x+1)^3$ .} \\ \text{Set:}\\ \qquad P(x)=b\cdot(x+1)^3 \\ \qquad \text{The root is $x = -1$ }\\ \text{so}\\ \qquad P(-1)=b(-1+1)^3=0 \\ \text{Set also:}\\ \qquad \text{ $P'(x)= 3b(x+1)^2 $}\\ \text{so}\\ \qquad P'(-1)= 3b(-1+1)^2=0 \\ \text{Set finally:}\\ \qquad \text{ $P''(x)= 6b(x+1) $}\\ \text{so}\\ \qquad P''(-1)= 6b(-1+1)^2=0 \\\\ \text{Conclusion} \\ \text{If $P(x)$ is divisible by $(x+1)^3$, so $\boxed{\mathbf{P(-1)=P'(-1)=P''(-1)=0}\qquad (2)}$ at the root $x=-1$ } \\ \end{array}\)

\(\begin{array}{lrcll} \text{We set:}\\ \text{1)}& f(x) + 2 &=& P(x) \quad & | \quad \text{$f(x)+2$ and $P(x)$ are divisible by $(x+1)^3 $ }\\ & f(-1) +2 &=& P(-1) = 0 \quad & | \quad \rightarrow (2) \\\\ \text{2)}& (f(x) + 2)' &=& P'(x) \\ & f'(x) &=& P'(x) \\ & f'(-1) &=& P'(-1) = 0 \quad & | \quad \rightarrow (2) \\\\ \text{3)}& (f(x) + 2)'' &=& P''(x) \\ & f''(x) &=& P''(x) \\ & f''(-1) &=& P''(-1) = 0 \quad & | \quad \rightarrow (2) \\ \end{array}\)

\(\begin{array}{lrcll} \text{We see:}\\ & \boxed{ \begin{array}{r} f(-1) +2 = 0 \qquad (3) \\ f'(-1) = 0 \qquad (4) \\ f''(-1)= 0 \qquad (5) \\ \end{array} } \end{array}\)

\(\begin{array}{lrcll} \text{We calculate:}\\ & f(x) &=& Ax^5+Bx^4+Cx^3 \\ & f(x) +2 &=& Ax^5+Bx^4+Cx^3 +2 \\ & f(-1)+2 &=& -A+B-C +2 = 0 \quad & | \quad \rightarrow (3) \\\\ & f'(x) &=& 5Ax^4+4Bx^3+3Cx^2 \\ & f'(-1) &=& 5A - 4B+3C = 0 \quad & | \quad \rightarrow (4) \\\\ & f''(x) &=& 20Ax^3+12Bx^2+6Cx \\ & f''(-1) &=& -20A+12B-6C = 0 \quad & | \quad \rightarrow (5) \\ \end{array}\)

\(\begin{array}{lrcll} \text{Solve the Simultaneous Equations, we calculate $A~$, $B~$, $C~$ :} \\ \end{array}\\ \begin{array}{lrcll} -A+B-C+2 = 0 \quad \Rightarrow & -A+B-C &=& -2 \\ & 5A-4B+3C &=& 0 \\ & -20A+12B-6C &=& 0 \\ \end{array}\)

Cramer's Rule:

\(\begin{array}{rcrcrcr} -1\cdot A &+& 1\cdot B &-& 1\cdot C &=& -2 \\ 5\cdot A &-& 4\cdot B &+& 3\cdot C &=& 0 \\ -20\cdot A &+& 12\cdot B &-& 6\cdot C &=& 0 \\ \end{array}\\ \small{ \begin{array}{lcl} \\ \text{Determinant denominator} &=& \begin{vmatrix} -1 & 1 & -1 \\ 5 & -4 & 3 \\ -20 & 12 & -6 \\ \end{vmatrix}\\ \\ &=& (-1)\cdot (-4)\cdot (-6) + 5\cdot 12\cdot (-1) +(-20)\cdot 1\cdot 3 - (-20)\cdot (-4)\cdot(-1) -(-1)\cdot 12\cdot 3 -5\cdot 1\cdot (-6) \\ &=& -24-60-60+80+36+30 \\ &=& 2 \\ \end{array} }\)

\(\small{ \begin{array}{lcl} A &=& \dfrac{ \begin{vmatrix} -2 & 1 & -1 \\ 0 & -4 & 3 \\ 0 & 12 & -6 \\ \end{vmatrix} }{2}\\\\ &=&\dfrac{ (-2)\cdot (-4)\cdot (-6) - (-2)\cdot 12\cdot 3 } {2}\\ &=&\dfrac{ 24 } {2}\\\\ \mathbf{A} & \mathbf{=} & \mathbf{12}\\ \end{array} \begin{array}{lcl} B &=& \dfrac{ \begin{vmatrix} -1 & -2 & -1 \\ 5 & 0 & 3 \\ -20 & 0 & -6 \\ \end{vmatrix} }{2}\\\\ &=&\dfrac{ (-20)\cdot (-2)\cdot 3 - 5\cdot(-2)\cdot (-6) } {2}\\ &=&\dfrac{ 60 } {2}\\\\ \mathbf{B} & \mathbf{=} & \mathbf{30}\\ \end{array} \begin{array}{lcl} C &=& \dfrac{ \begin{vmatrix} -1 & 1 & -2 \\ 5 & -4 & 0 \\ -20 & 12 & 0 \\ \end{vmatrix} }{2}\\\\ &=&\dfrac{ 5\cdot 12\cdot (-2) - (-20)\cdot (-4)\cdot (-2) } {2}\\ &=&\dfrac{ 40 } {2}\\\\ \mathbf{C} & \mathbf{=} & \mathbf{20}\\ \end{array} }\)

\(\begin{array}{lrcll} \text{The Polynom $f(x) = Ax^5+Bx^4+Cx^3$ with $A=12$, $~B=30$, and $~C = 20$}\\ \quad \boxed{f(x)=12x^5+30x^4+20x^3} \end{array}\)

Proof:

\(\begin{array}{|rcll|} \hline f(3) &=& 12\cdot 3^5 + 30\cdot 3^4 + 20\cdot 3^3 \\ &=& 2916 + 2430 + 540 \\ &=& 5886 \\\\ \frac{f(3)}{3^3} &=& \frac{5886}{3^3} \\\\ &=& 218\ \checkmark \\\\ \frac{f(3)+2}{(3+1)^3} &=& \frac{5886+2}{4^3} \\\\ &=& \frac{5888}{64} \\\\ &=& 92\ \checkmark \\ \hline \end{array}\)

Thanks Loki and Heureka,

I went with a more straight forward (but not shorter) approach, 

I just divided by (x+1) three times to find the coefficients.

\(f(x)=ax^5+bx^4+cx^3\\ f(x)+2=ax^5+bx^4+cx^3+2\\ [f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\ \text{I did this by algebraic division and got a remainer of }-c+b-a+2\\ \text{The remainder must be 0 so}\\ -c+b-a+2=0\\ c=b-a+2 \)

So now I have

\(f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\ [f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\ \qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }6-b+2a\\ \text{The remainder must be 0 so}\\ 6-b+2a=0\\ b=6+2a\\~\\ (ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\ =(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\ =ax^3+6x^2-4x+2 \)

\((ax^3+6x^2-4x+2)\div(x+1)\\ \\\text{I did this by algebraic division and got a remainder of }12-a\\ \text{The remainder must be 0 so}\\ a=12\\~\\ so\\ a=12\\ b=6+2a=30\\ c=b-a+2=30-12+2=20\\ so\\ f(x)=12x^5+30x^4+20x^3\)

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LaTex:

f(x)=ax^5+bx^4+cx^3\\
f(x)+2=ax^5+bx^4+cx^3+2\\
[f(x)+2]\div(x+1)=[ax^5+bx^4+cx^3+2]\div(x+1)\\
\text{I did this by algebraic division and got a remainer of }-c+b-a+2\\
\text{The remainder must be 0 so}\\
-c+b-a+2=0\\
c=b-a+2

f(x)+2=ax^5+bx^4+(b-a+2)x^3+2\\
[f(x)+2]\div(x+1)\\\qquad=[ax^5+bx^4+(b-a+2)x^3+2]\div(x+1)\\
\qquad =ax^4+(b-a)x^3+2x^2-2x+2\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }6-b+2a\\
\text{The remainder must be 0 so}\\
6-b+2a=0\\
b=6+2a\\~\\
(ax^4+(b-a)x^3+2x^2-2x+2)\div(x+1)\\
=(ax^4+(6+a)x^3+2x^2-2x+2)\div(x+1)\\
=ax^3+6x^2-4x+2

(ax^3+6x^2-4x+2)\div(x+1)\\
\\\text{I did this by algebraic division and got a remainder of }12-a\\
\text{The remainder must be 0 so}\\
a=12\\~\\
so\\
a=12\\
b=6+2a=30\\
c=b-a+2=30-12+2=20\\
so\\
f(x)=12x^5+30x^4+20x^3