+0  
 
+1
680
1
avatar+598 

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

 $f(x)-1$ is divisible by $(x-1)^3$.

 $f(x)$ is divisible by $x^3$.

michaelcai  Jan 23, 2018
 #1
avatar+20632 
0

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)-1$ is divisible by $(x-1)^3$.
$f(x)$ is divisible by $x^3$.


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

 

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

 

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

 

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

 

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

 

\(\begin{array}{lrcll} \text{Solve the Simultaneous Equations, we calculate $A~$, $B~$, $C~$ :} \\ \end{array}\\ \begin{array}{lrcll} A+B+C -1 = 0 \quad \Rightarrow & A+B+C &=& 1 \\ & 5A+4B+3C &=& 0 \\ & 20A+12B+6C &=& 0 \\ \end{array}\)

 

Cramer's Rule:

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

 

\(\begin{array}{lcl} A &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 0&4&3 \\ 0&12&6 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 1\cdot 4\cdot 6 - 1\cdot 12\cdot 3 } {2}\\ &=&\dfrac{ -12 } {-2}\\\\ \mathbf{A} & \mathbf{=} & \mathbf{6}\\ \end{array} \begin{array}{lcl} B &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 5&0&3 \\ 20&0&6 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 20\cdot 1\cdot 3 - 5\cdot 1\cdot 6 } {-2}\\ &=&\dfrac{ 30 } {-2}\\\\ \mathbf{B} & \mathbf{=} & \mathbf{-15}\\ \end{array} \begin{array}{lcl} C &=& \dfrac{ \begin{vmatrix} 1&1&1 \\ 5&4&0 \\ 20&12&0 \\ \end{vmatrix} }{-2}\\\\ &=&\dfrac{ 5\cdot 12\cdot 1 - 20\cdot 4\cdot 1 } {-2}\\ &=&\dfrac{ -20 } {-2}\\\\ \mathbf{C} & \mathbf{=} & \mathbf{10}\\ \end{array}\)

 

\(\begin{array}{lrcll} \text{The Polynom $f(x) = Ax^5+Bx^4+Cx^3$ with $A=6$, $~B=-15$, and $~C = 10$}\\ \quad \boxed{f(x)=6x^5-15x^4+10x^3} \end{array}\)

 

Proof:

\(\begin{array}{|rcll|} \hline f(3) &=& 6\cdot 3^5-15\cdot 3^4 + 10 \cdot 3^3 \\ &=& 1458-1215+270 \\ &=& 513 \\ \frac{f(3)}{3^3} &=& \frac{513}{3^3} \\ &=& 19\ \checkmark \\ \frac{f(3)-1}{(3-1)^3} &=& \frac{513-1}{2^3} \\ &=& \frac{514}{8} \\ &=& 64\ \checkmark \\ \hline \end{array}\)

 

 

laugh

heureka  Jan 25, 2018

10 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.