2, -2-5i; degree 3

Question

I need to know how to solve a polynomial f(x) with real coefficients and a leading coefficient 1 with the given zeros and degree. 

Answer 1

 

 

Answer 2

2 is a zero so (x-2) is a factor

-2-5i is a zero and therefore its conjugate is a zero, that is -2+5i

[x-(-2-5i)][x-(-2+5i)] are factors

[x+2+5i][x+2-5i]

rewrite these as [(x+2)+5i] and [(x+2)-5i]

multiply these together keeping in mind you have the difference of two squares

[(x+2)+5i][(x+2)-5i]=...

(x+2)²-(5i)²=...

(x²+4x+4-(25i²) where i²=-1

(x²+4x+4-(-25))=...

(x²+4x+4+25)=...

(x²+4x+29) is another factor

multiply these together

(x-2)(x²+4x+29)=x³+4x²+29x-2x²-8x-58=x³+2x²+21x-58=0 is the polynomial equation

the difficult part is writing the factors with the complex zeroes, keeping in mind that a complex zero has a conjugate