How do you write a polynomial with zeros -2, -2, 3, -4i and leading coefficient 1?

Question

1546 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T12:40:52

# (x+2)^2(x-3)(x^2+16) #

We want the following roots:

# -2, -2, 3, -4i #

Complex roots appear in conjugate pairs so the complex root #-4i# must also be paired with a complex root #+4i#, so we require the following roots:

# -2, -2, 3, 4i, -4i #

By the factor theorem if #x=alpha# is a root of #f(x)=0#, then #(x-alpha)# is a factor of #f(x)#.

For the roots #-2,-2# (double root) we require two factors:

# (x+2)(x+2) = (x+2)^2#

For the root #3# we require the factor:

# (x-3) #

For the roots #+-4i# we require the (complex) factors;

# (x-4i)(x+4i) = x^2+16 #

Combining these results we get:

# f(x) = (x+2)^2(x-3)(x^2+16) #