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Aus Online Mathematik Brückenkurs 2

First, we try to determine the pure imaginary root.

We can write the imaginary root as z=ia, where a is a real number. If we substitute in z=ia, the equation should then be satisfied,

(ia)4+3(ia)3+(ia)2+18(ia)−30=0

i.e.

a4−3a3i−a2+18ai−30=0

and, if collect together the real and imaginary parts on the left-hand side, we have

(a4−a2−30)+a(−3a2+18)i=0.

If both sides are to be equal, the left-hand side's real and imaginary parts must be zero,

a4−a2−30a(−3a2+18)=0=0.

The other relation gives a=0 or a=6 , but it is only a=6  which satisfies the first relation.

Thus, the equation z4+3z3+z2+18z−30=0 has two pure imaginary roots, z=−i6  and z=i6 . Note that it is completely normal to obtain two imaginary roots. The polynomial equation has real coefficients and must therefore have complex conjugate roots.

Now we tackle the problem of determining the equation's other two roots. Because we know that the equation has the two roots z=i6 , the factor theorem gives that the equation contains the factor

(z−i6)(z+i6)=z2+6

i.e. we can factorize the left-hand side of the equation in the following way,

z4+3z3+z2+18z−30=(z2+Az+B)(z2+6)

where the equation's two other roots are zeros of the unknown factor z2+Az+B.

We determine the factor z2+Az+B by means of a polynomial division (divide both sides by z2+6),

z2+Az+B=z2+6z4+3z3+z2+18z−30=z2+6z4+6z2−6z2+3z3+z2+18z−30=z2+6z2(z2+6)+3z3−5z2+18z−30=z2+z2+63z3−5z2+18z−30=z2+z2+63z3+18z−18z−5z2+18z−30=z2+z2+63z(z2+6)−5z2−30=z2+3z+z2+6−5z2−30=z2+3z+z2+6−5(z2+6)=z2+3z−5.

To obtain the two remaining roots, we need therefore to solve the equation

z2+3z−5=0.

We complete the square

z+232−232−5z+232=0=429

which gives that z=−23229 .

The answer is that the equation has the roots

z=−i6 , z=i6 , z=−23−229 , z=−23+229.