Polynomial Division: dividing by a double root

divisibilitypolynomials

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.

Best Answer

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...