1+i is a zero of f (x) = x^4-2x^3-x^2+6x-6

Question

Asa Rivas

Mathematics

17 February, 14:07

1+i is a zero of f (x) = x^4-2x^3-x^2+6x-6

+4

Answers (1)

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Kadyn Mcbride · Answer 1

F (1+i) = (1+i) ^4 - 2 (1+i) ^3 - (1+i) ^2 + 6 (1+i) - 6;

But, (1+i) ^2 = 1^2 + 2*1*i + i^2 = 1 + 2i - 1 = 2i; because i^2 = - 1;

(1+i) ^4 = [ (1+i) ^2]^2 = (2i) ^2 = 4*i^2 = - 4;

(1+i) ^3 = (1+i) ^2 * (1+i) = 2i * (1 + i) = 2i + 2i^2 = 2i - 2;

Then, f (1+i) = - 4 - 2 (2i - 2) - 2i + 6 (1 + i) - 6 = - 4 + 4i + 4 + 6 + 6i - 6 = 10i which isn't 0;

Finally, 1+i isn't zero of f (x);