Polynomial function

Question

find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and roots of 2-square root of 6, 2+square root of 6, and 3-i

Answer 1

roots 2-sqr6, 2+sqr6, 3-i, 3+i

imaginary roots come in conjugate pairs

factors are (x-2-sqr6)(x-2+sqr6)(x-3-i)(x-3+i)

= (x^2-6x+10)(x^2-4x-2)

= (x^4 -10x^3 +32x^2 -28x -20

Answer 2

desmos.com/calculator/lmjcnn6cso

Answer 3

Hi,

roots of 2-square root of 6, 2+square root of 6, and 3-i

This expression it means: all roots are 2 - √6 2+√6 3-i 3+i

point: when one of root is a + ib you can be sure another root is a - ib

ask you :least degree so this polynomial is at least 4

p = ( x - (2 - √6) ) ( x - (2 + √6) ) ( x - (3 - i) ) (x - (3 +i) )

p = ( x - 2 + √6) ( x - 2 - √6) ( x - 3 + i) (x - 3 -i)

p = ( x^2 -4x -2) ( x^2 -6x +10) After distribute you have a polynomial that its degree is 4

p = x^4 -10 x^3 +32 x^2 -28 x -20

I hope it is useful,

Minoo