Respuesta :
Polynomials
A polynomial can be built if we know its zeros, also called roots
Suppose p,q, and r are the roots of a polynomial of degree 3, then:
p(x)=a(x-p)(x-q)(x-r)
Where a is a real number different from 0
Note we are only given two roots:
p=1
q=i
recall that i is the base of the complex numbers, that is:
[tex]\mathbf{i}=\sqrt{-1}[/tex]The third root comes when we recall that, if a polynomial has real coefficients, the complex roots come in conjugate pairs, i.e. if a+bi is one root of the polynomial, then a-bi is also a root of the polynomial.
Thus, the other root is the conjugate of q:
r=-i
Now we have all the roots, we just apply the above equation to find:
p(x)=a(x-p)(x-q)(x-r)
p(x)=a(x-1)(x-i)(x+i)
Recall that
[tex]\mleft(x-i\mright)\mleft(x+i\mright)=x^2-i^2[/tex]Since
[tex]i^2=-1[/tex][tex](x-i)(x+i)=x^2+1[/tex]Finally, the required polynomial is:
[tex]p\mleft(x\mright)=a\mleft(x-1\mright)(x^2+1)[/tex]Since no other condition is given, we choose a=1:
[tex]p(x)=(x-1)(x^2+1)[/tex]This is the required polynomial
Operating the products:
[tex]p(x)=(x-1)(x^2+1)=x^3+x-x^2-1[/tex]Ordering:
[tex]p(x)=x^3-x^2+x-1[/tex]