Question

how do you use the given zero to fnd the remaining zeros of the function h(x)=4x^4+7x^3+62x^2+112x-32; zero:-4i
help please!!! I am lost

Answer 1

if -4i is solution, then 4i is also a solution, so you can factor H(x) as follows:

(x+4i)(x-4i)(2nd degree polynomial).

Now, (x+4i)(x-4i)=x²+16, so do a long division: (using European notation, sorry)

x²+16 / 4x^4+7x^3+62x^2+112x-32 \ 4x² + 7x - 2
4x^4 +64x^2
------------------ -
7x^3 - 2x^2
7x^3 +112x
----------------- -
-2x^2 - 32
-2x^2 - 32
--------------- -
0

So it's factored as:

(x+4i)(x-4i)(4x²+7x-2)

Factor the 2nd degree part to find the two real zeroes!