I have this fairly interesting problem that is based on Polynomial Division and/or factor/remainder theorem.

Determine the values of $a$ and $b$ such that $ax^4 + bx^3 -3$ is divisible by $(x-1)^2$.

This is interesting because the root we are dividing by is a double root, so its difficult to get $2$ equations and $2$ unknowns. (note that one cannot use ideas from Calculus)

The only approach I have been able to come up with is to expand the perfect square divisor to a full quadratic and perform a brute force division, but it did not really lead me to a solution.

  • 3
    $\begingroup$ What was your approach so far? $\endgroup$ – Jonas Aug 4 '18 at 17:14

Method one (compare the coefficients): $$\begin{eqnarray}ax^4+bx^3-3 &=& (x-1)^2(cx^2+dx+e)\\ &=& (x^2-2x+1)(cx^2+dx+e)\\ &=& cx^4+(d-2c)x^3+(e-2d+c)x^2+(-2e+d)x+e\end{eqnarray}$$ From here we see that:
$$ \begin{eqnarray*} c&=&a\\ e&=& -3\\ -2e+d&=& 0\implies d=-6\\ e-2d+c&=&0 \implies c=-9\implies a=-9\\ d-2c&=&b\implies b=-12 \end{eqnarray*}$$

Method two (Vieta formulas), $x_1=x_2=1$: $$ 2+x_3+x_4 = -{b\over a}$$ $$ 1+2(x_3+x_4)+x_3x_4 = 0$$ $$ 2x_3x_4+ x_3+ x_4 =0$$ $$ x_3x_4 = -{3\over a}$$ from 2. and 3. equation we get $$x_3+x_4 = -{2\over 3}\;\;\;{\rm and}\;\;\;x_3x_4 ={1\over 3}$$ From 4. equation we get $a=-9$ and from 1. equation we get $b=-12$.

Method three Since $1$ is root of a polynomial $p(x)=ax^4+bx^3-3$ we get $b=3-a$ so we have $$p(x)=ax^4+3x^3 -ax^3-3 $$ $$= ax^3(x-1)+3(x-1)(x^2+x+1) $$ $$= (x-1)\underbrace{\Big(ax^3+3(x^2+x+1)\Big)}_{q(x)} $$ Now since $1$ double root we have also $q(1)=0$ so $a+9=0$.

Method four: (Horner schema) $$\begin{array}{cccccc} & a & b & 0 & 0 & -3 \\ \hline 1 & & a & a+b & a+b & a+b \\ \hline & a & a+b & a+b & a+b & \color{red}{a+b-3} \\ \hline 1 & & a & 2a+b & 3a+2b & \\ \hline & a & 2a+b & 3a+2b & \color{red}{4a+3b} & \\ \end{array}$$ Both red expressions must be zero...

Method five: Direct (long) division. What you are left ($1$. degree polynomial ) must be identical to $0$, so you get $2$ equations again...

  • $\begingroup$ Hi Greediod, i will award you to be the solution. Because you put so much effort and did it multiple ways. You are like the mathematician Karl Friedrich Gauss, the way he gave 10 different proofs for a particular Number Theory problem. Thanks again. $\endgroup$ – Palu Aug 5 '18 at 3:55

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