Question

Write a third degree polynomial function y = P(x) with rational coefficients so that P(x) = 0 has roots -2 and 6 + i

Answer 1

Answer:$P(x)=x^3-10x^2+13x+74$

Step-by-step explanation: We are given roots of a third degree polynomial function -2, 6+i.

Note: a radical always comes with pair of plus and minus sign.

Therefore, there would be one more root 6-i.

So, all the roots would be -2, 6+i, 6-i.

So, the factors of polynomial function would be

(x+2)(x-6-i)(x-6+i)

Multiplying those factors,

$(x+2)(x^2-6x+xi -6x+36-6i-ix+6i-i^2)$

$(x+2)(x^2-12x+36+1)$

$(x+2)(x^2-12x+37)$

$x^3-12x^2+2x^2+37x-24x+74$

$=x^3-10x^2+13x+74$

Therefore,

$P(x)=x^3-10x^2+13x+74$