Use the given zero to find all the zeros of the function, Please explain I do not understand?

Given zero: 3i

(The i is imaginary)

Function: f(x)=x³+x²+9x+9

1 Answer
Dec 4, 2016

Answer:

#{-3i, 3i, -1}#

Explanation:

The zeros, or roots, of a function #f(x)# are the solutions to the equation #f(x) = 0#. I.e. they are the values which return #0#, when entered into #f(x)#.

This problem uses two properties of roots.

  • #x_0# is a zero of a polynomial #P(x)# if and only if #(x-x_0)# is a factor of #P(x)#.

  • If #P(x)# is a polynomial with real coefficients and #z=a+bi# is a nonreal complex number which is a zero of #P(x)#, then its complex conjugate #bar(z)=a-bi# is also a zero of #P(x)#.

As #f(x) = x^3+x^2+9x+9# is a polynomial with real coefficients, and #3i = 0+3i# is a zero of #f(x)#, then the second property gives us that #0-3i=-3i# must also be a zero of #f(x)#.

Because #3i# and #-3i# are zeros of #f(x)#, the first property gives us that #(x+3i)# and #(x-3i)# are both factors of #f(x)#. Thus their product #(x+3i)(x-3i) = x^2+9# is also a factor of #f(x)#.

As #f(x)# is a degree #3# polynomial, it must have exactly #3# zeros (with the possibility of some repeating). We have two of them, so let's let #x_0# be the third zero. Then we know #(x-x_0)# is a factor of #f(x)#, and it is indeed the last nonconstant factor, so for some nonzero #c#,

#x^3+x^2+9x+9= c(x+3i)(x-3i)(x-x_0)#

#=c(x^2+9)(x-x_0)#

#=cx^3-cx_0x^2+9cx-9cx_0#

Finally, we equate coefficients to find the unknown values. Equating the coefficients of the #x^3# terms, we have #1=c#. Equating the coefficients of the constant terms, we have #9=-9cx_0 = -9x_0#. Thus #x_0 = -1#

So, all together, #f(x)# has the zeros #{-3i, 3i, -1}#