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Aus Online Mathematik Brückenkurs 2

A polynomial is said to have a triple root z=c if the equation contains the factor z−c3 

For our equation, this means that the left-hand side can be factorized as

z4−6z2+az+b=z−c3z−d 

according to the factor theorem, where z=c is the triple root and z=d is the equation's fourth root (according to the fundamental theorem of algebra, a fourth-order equation always has four roots, taking into account multiplicity).

We will now try to determine a, b, c and d so that both sides in the factorization above agree.

If we expand the right-hand side above, we get

z−c3z−d=z−c2z−cz−d=z2−2cz+c2z−cz−d=z3−3cz2+3c2z−c3z−d=z4−3c+dz3+3cc+dz2−c2c−3dz+c3d

and this means that we must have

z4−6z2+az+b=z4−3c+dz3+3cc+dz2−c2c−3dz+c3d 

Because two polynomials are equal if an only if their coefficients are equal, this gives

3c+d=03cc+d=−6−c2c−3d=ac3d=b

From the first equation, we obtain d=−3c and substituting this into the second equation gives us an equation for c,

3cc−3c=−6−6c2=−6 

i.e. c=−1 or c=1. The relation d=−3c gives that the corresponding values for d are d=3 and d=−3. The two last equations give us the corresponding values for a and b,

\displaystyle \begin{align} & c=1,\ d=-3:\quad a=-1^{2}\centerdot \left( 1-3\centerdot \left( -3 \right) \right)=8 \\ & b=1^{3}\centerdot \left( -3 \right)=-3 \\ \end{align}

\displaystyle \begin{align} & c=-1,\ d=3:\quad a=-\left( -1 \right)^{2}\centerdot \left( -1-3\centerdot 3 \right)=10 \\ & b=\left( -1 \right)^{3}\centerdot 3=-3 \\ \end{align}

Therefore, there are two different answers:

\displaystyle \bullet \quad a=\text{8 } and \displaystyle b=-\text{3} give the triple root \displaystyle z=\text{1} and the single root \displaystyle z=-\text{3};

\displaystyle \bullet \quad a=10 and \displaystyle b=-\text{3 }give the triple root \displaystyle z=-\text{1 } and the single root \displaystyle z=\text{3}.