Question

Write a polynomial function with rational coefficients so that P(x) = 0 has the given roots. 4, 16, and 1 + 9i

Answer 1

I remember dis, yey

so if a polynomial has roots $r_1$, $r_2$, $r_3$, it can be factored into
$f(x)=a(x-r_1)^b(x-r_2)^c(x-r_3)^d$ where a,b,c,d are constants

also, if a polynomial has rational coefients and a+bi is a root, then a-bi must also be a root


so our roots we need are
4,16, 1+9i and 1-9i

so assuming multiplity 1 (that means we have something like [/tex]f(x)=a(x-r_1)^1(x-r_2)^1(x-r_3)^1[/tex])

we get that your function is
$P(x)=(x-4)(x-16)(x-(1+9i))(x-(1-9i))$ which simplifies to
$P(x)=(x-4)(x-16)(x-1-9i)(x-1+9i)$ which expands to
$P(x)=x^4-22x^3+186x^2-1768x+5248$