Question

. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i

Answer 1

Answer:

Error of Andrew: Made incorrect factors from the roots

Step-by-step explanation:

Roots of the polynomial are: 3, 2 + 2i, 2 - 2i. According to the factor theorem, if a is a root of the polynomial P(x), then (x - a) is  a factor of P(x). According to this definition:

(x - 3) , (x - (2 + 2i)) , (x - (2 - 2i)) are factors of the required polynomial.

Simplifying the brackets, we get:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) are factors of the required polynomial.

This is the step where Andrew made the error. The factors will always be of the form (x - a) , not (x + a). Andrew wrote the complex factors in form of (x + a)  which resulted in the wrong answer.

So, the polynomial would be:

$(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0$