Step 1

Given: \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}\)

It is given that 4i is a zero of P(x)

We know that for polynomials with real coefficients, the complex zeros always exists in conjugate pairs

Hence, -4i is also a zero of P(x)

Hence, \(\displaystyle{\left({x}-{4}{i}\right)},{\left({x}+{4}{i}\right)}\) are factors of P(x)

So, \(\displaystyle{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}\) is a factor of P(x)

\(\displaystyle\Rightarrow{x}^{{{2}}}+{16}\) is a factor of P(x)

Dividing P(x) by \(\displaystyle{x}^{{{2}}}+{16}\)

So, \(\displaystyle{\frac{{{3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}}}{{{x}^{{{2}}}+{16}}}}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)

\(\displaystyle{P}{\left({x}\right)}={\left({3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\right)}{\left({x}^{{{2}}}+{16}\right)}\)

Step 2

Consider \(\displaystyle{F}{\left({x}\right)}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)

We observe that \(\displaystyle{F}{\left(-{2}\right)}={3}{\left(-{2}\right)}^{{{3}}}-{8}{\left(-{2}\right)}^{{{2}}}-{13}{\left(-{2}\right)}+{30}\)

\(\displaystyle{F}{\left(-{2}\right)}=-{24}-{32}+{26}+{30}\)

\(\displaystyle{F}{\left(-{2}\right)}={0}\)

Similarly, \(\displaystyle{F}{\left({3}\right)}={0}\)

So, -2,3 are zeros of F(x) and hence of P(x) too

Let \(\displaystyle\alpha\) be the fifth zero of P(x)

We know that the sum of zeros of \(\displaystyle{P}{\left({x}\right)}=\frac{{8}}{{3}}\)

\(\displaystyle\alpha+{4}{i}-{4}{i}+{\left(-{2}\right)}+{\left({3}\right)}={\frac{{{8}}}{{{3}}}}\)

\(\displaystyle\alpha={\frac{{{8}}}{{{3}}}}-{1}\)

\(\displaystyle\alpha={\frac{{{5}}}{{{3}}}}\)

Hence, the zeros are \(\displaystyle-{2},{3},{4}{i},-{4}{i},{\frac{{{5}}}{{{3}}}}\)

Hence, \(\displaystyle{P}{\left({x}\right)}={3}{\left({x}+{2}\right)}{\left({x}-{3}\right)}{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}{\left({x}-{\frac{{{5}}}{{{3}}}}\right)}\)

Given: \(\displaystyle{P}{\left({x}\right)}={3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}\)

It is given that 4i is a zero of P(x)

We know that for polynomials with real coefficients, the complex zeros always exists in conjugate pairs

Hence, -4i is also a zero of P(x)

Hence, \(\displaystyle{\left({x}-{4}{i}\right)},{\left({x}+{4}{i}\right)}\) are factors of P(x)

So, \(\displaystyle{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}\) is a factor of P(x)

\(\displaystyle\Rightarrow{x}^{{{2}}}+{16}\) is a factor of P(x)

Dividing P(x) by \(\displaystyle{x}^{{{2}}}+{16}\)

So, \(\displaystyle{\frac{{{3}{x}^{{{5}}}-{8}{x}^{{{4}}}+{35}{x}^{{{3}}}-{98}{x}^{{{2}}}-{208}{x}+{480}}}{{{x}^{{{2}}}+{16}}}}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)

\(\displaystyle{P}{\left({x}\right)}={\left({3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\right)}{\left({x}^{{{2}}}+{16}\right)}\)

Step 2

Consider \(\displaystyle{F}{\left({x}\right)}={3}{x}^{{{3}}}-{8}{x}^{{{2}}}-{13}{x}+{30}\)

We observe that \(\displaystyle{F}{\left(-{2}\right)}={3}{\left(-{2}\right)}^{{{3}}}-{8}{\left(-{2}\right)}^{{{2}}}-{13}{\left(-{2}\right)}+{30}\)

\(\displaystyle{F}{\left(-{2}\right)}=-{24}-{32}+{26}+{30}\)

\(\displaystyle{F}{\left(-{2}\right)}={0}\)

Similarly, \(\displaystyle{F}{\left({3}\right)}={0}\)

So, -2,3 are zeros of F(x) and hence of P(x) too

Let \(\displaystyle\alpha\) be the fifth zero of P(x)

We know that the sum of zeros of \(\displaystyle{P}{\left({x}\right)}=\frac{{8}}{{3}}\)

\(\displaystyle\alpha+{4}{i}-{4}{i}+{\left(-{2}\right)}+{\left({3}\right)}={\frac{{{8}}}{{{3}}}}\)

\(\displaystyle\alpha={\frac{{{8}}}{{{3}}}}-{1}\)

\(\displaystyle\alpha={\frac{{{5}}}{{{3}}}}\)

Hence, the zeros are \(\displaystyle-{2},{3},{4}{i},-{4}{i},{\frac{{{5}}}{{{3}}}}\)

Hence, \(\displaystyle{P}{\left({x}\right)}={3}{\left({x}+{2}\right)}{\left({x}-{3}\right)}{\left({x}-{4}{i}\right)}{\left({x}+{4}{i}\right)}{\left({x}-{\frac{{{5}}}{{{3}}}}\right)}\)