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h (x)=3x^4+13x^3+2x^2+52x-40, zero:-2i
the remaining zeros of h are?

h(x)=3x^4 +13x^3 +2x^2 +52x -40

complex zeroes always occurs in pair

SO if -2i is a zero,then 2i is also a zero

Therefore in factor form,we can write it as

(x-2i) , (x+2i)

Next step is to reduce it

And for that we have to divide the given polynomial by (x-2i)(x+2i)

And on dividing we will get 3x^2 +13x-10 which can be factored into (x+5)(3x-2)

To find the zero we have to set them to zero

(x+5)(3x-2)=0

x=-5,2/3

Therefore the zeroes are 2i,-2i,-5,2/3