farewell
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Good evening everyone,
here is the statement:
Note the equation (E):
Z³-12z² + = 48z-128.
1.Check that 8 is the solution of (E)
2.a) determine real a, b, c such that for all z C (overall)
Z³ 12z²-48z-128 + = (z-8) (az² + bz + c).
b) Solve the equation in C (E)
For 1 I thought about putting (z-8) factor which gives:
(z-8) z (z²-12z + 48) -128√?
after I have a polynomial of degree 2 and calculating the discriminant I find z1: 6-i√12 and z2: 6 + i√12
and there I do not know how ... can you enlighten me
here is the statement:
Note the equation (E):
Z³-12z² + = 48z-128.
1.Check that 8 is the solution of (E)
2.a) determine real a, b, c such that for all z C (overall)
Z³ 12z²-48z-128 + = (z-8) (az² + bz + c).
b) Solve the equation in C (E)
For 1 I thought about putting (z-8) factor which gives:
(z-8) z (z²-12z + 48) -128√?
after I have a polynomial of degree 2 and calculating the discriminant I find z1: 6-i√12 and z2: 6 + i√12
and there I do not know how ... can you enlighten me