Step 1

Given \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}+{5}{x}^{{{2}}}+{25}{x}+{125}\)

We have to write P in factored from

Step 2

given: \(\displaystyle{p}{\left({x}\right)}={x}^{{{3}}}+{5}{x}^{{{2}}}+{25}{x}+{125}\)

\(\displaystyle{p}{\left({x}\right)}={\left({x}^{{{3}}}+{5}{x}^{{{2}}}\right)}+{\left({25}{x}+{125}\right)}\)

\(\displaystyle{p}{\left({x}\right)}={x}^{{{2}}}{\left({x}+{5}\right)}+{25}{\left({x}+{5}\right)}\)

\(\displaystyle{p}{\left({x}\right)}={\left({x}^{{{2}}}+{25}\right)}{\left({x}+{5}\right)}\)

Given \(\displaystyle{P}{\left({x}\right)}={x}^{{{3}}}+{5}{x}^{{{2}}}+{25}{x}+{125}\)

We have to write P in factored from

Step 2

given: \(\displaystyle{p}{\left({x}\right)}={x}^{{{3}}}+{5}{x}^{{{2}}}+{25}{x}+{125}\)

\(\displaystyle{p}{\left({x}\right)}={\left({x}^{{{3}}}+{5}{x}^{{{2}}}\right)}+{\left({25}{x}+{125}\right)}\)

\(\displaystyle{p}{\left({x}\right)}={x}^{{{2}}}{\left({x}+{5}\right)}+{25}{\left({x}+{5}\right)}\)

\(\displaystyle{p}{\left({x}\right)}={\left({x}^{{{2}}}+{25}\right)}{\left({x}+{5}\right)}\)