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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−c)3.

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

z4−6z2+az+b=(z−c)3(z−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−c)3(z−d)=(z−c)2(z−c)(z−d)=(z2−2cz+c2)(z−c)(z−d)=(z3−3cz2+3c2z−c3)(z−d)=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d

and this means that we must have

z4−6z2+az+b=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d.

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

3c+d3c(c+d)−c2(c−3d)c3d=0=−6=a=b.

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

3c(c−3c)−6c2=−6=−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,

c=1 d=−3:abc=−1 d=3:ab=−12(1−3(−3))=8=13(−3)=−3=−(−1)2(−1−33)=10=(−1)33=−3.

Therefore, there are two different answers,

• a=8 and b=−3 give the triple root z=1 and the single root z=−3,

• a=10 and b=−3 give the triple root z=−1 and the single root z=3.