Question

write a polynomial function of leaswt degree with integral coefficients that have the given zeros.: -2,5,-3i

Answer 1

(x+2)(x-5)(((-x)^2)+9)

Answer 2

solutions are given as -2,5,-3i,therefore
the polynomial equation will be

\(\large\tt \color{black}{=(x+2)(x-5)(x-3i)(x+3i)}\)

Answer 3

simplify it now

Answer 4

its very easy

Answer 5

first lets simplify (x+2)(x-5)

Answer 6

(x+2)(x-5)
=x^-5x+2x-10
=x^2-3x-10

Answer 7

now lets simplify
(x-3i)(x+3i)

u can simplify it as i did above
and use formula here for quick solution
(a+b)(a-b)=a^2-b^2

hence
(x-3i)(x+3i)
=x^-(3i)^2
=x^2-9i^2
=x^2+9..............(as i=\[ i=\sqrt{-1}\]
(i^2=-1)
now the equation will be
(x+2)(x-5)(x-3i)(x+3i)
=(x^2-3x-10)(x^2+9)
=x^4+9x^2-3x^3-27x-10x^2-90
=x^4-3x^3-x^2-27x-90
so this is ur polynomial function

f(x)==x^4-3x^3-x^2-27x-90

Answer 8

\(\huge\tt \color{black}{f(x)=x^4-3x^3-x^2-27x-90}\)