Step 1

Given:

f(x) is a polynomial with real coefficients of degree 3

Zeroes of f(x) are -4,2 + 3i

Since f(x) is a polynomial with real coefficients the complex roots exist as complex cojugates

So, 2-3i is a zero of f(x) as 2+3i is a zero.

Step 2

Hence, f(x)=(x-4)(x-(2+3i))(x-(2-3i))

f(x)=(x-(-4))((x-2)-3i)((x-2)+3i)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}+{4}\right)}{\left({\left({x}-{2}\right)}^{{2}}+{9}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}+{4}\right)}{\left({x}^{{2}}-{4}{x}+{13}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}-{3}{x}+{52}\)

Hence, the polynomial is \(\displaystyle{x}^{{3}}-{3}{x}+{52}\)

Given:

f(x) is a polynomial with real coefficients of degree 3

Zeroes of f(x) are -4,2 + 3i

Since f(x) is a polynomial with real coefficients the complex roots exist as complex cojugates

So, 2-3i is a zero of f(x) as 2+3i is a zero.

Step 2

Hence, f(x)=(x-4)(x-(2+3i))(x-(2-3i))

f(x)=(x-(-4))((x-2)-3i)((x-2)+3i)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}+{4}\right)}{\left({\left({x}-{2}\right)}^{{2}}+{9}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={\left({x}+{4}\right)}{\left({x}^{{2}}-{4}{x}+{13}\right)}\)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}-{3}{x}+{52}\)

Hence, the polynomial is \(\displaystyle{x}^{{3}}-{3}{x}+{52}\)