Step 1

Consider the given function as

\(\displaystyle{P}{\left({x}\right)}={12}{x}^{{{3}}}-{25}{x}^{{{2}}}+{x}+{2}\)

Step 2

Let x =2 is a zero

\(\displaystyle{P}{\left({2}\right)}={12}{\left({2}\right)}^{{{3}}}-{25}{\left({2}\right)}^{{{2}}}+{2}+{2}\)

=12(8)-25(4)+4

=96-100+4

=0

So, x=2 is a zero.

Now find the other zeros as

Now

\(\displaystyle{\frac{{{12}{x}^{{{3}}}-{25}{x}^{{{2}}}+{x}+{2}}}{{{x}-{2}}}}={\frac{{{\left({3}{x}-{1}\right)}{\left({4}{x}^{{{2}}}-{7}{x}-{2}\right)}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}^{{{2}}}-{\left({8}-{1}\right)}{x}-{2}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}^{{{2}}}-{8}{x}+{1}{x}-{2}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}{\left({x}-{2}\right)}+{1}{\left({x}-{2}\right)}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left({x}-{2}\right)}{\left({4}{x}+{1}\right)}}}{{{x}-{2}}}}\)

=(3x-1)(4x+1)

Step 3

Hence, the zero are x = 2, \(\displaystyle{\frac{{{1}}}{{{3}}}},{\frac{{-{1}}}{{{4}}}}\)

Consider the given function as

\(\displaystyle{P}{\left({x}\right)}={12}{x}^{{{3}}}-{25}{x}^{{{2}}}+{x}+{2}\)

Step 2

Let x =2 is a zero

\(\displaystyle{P}{\left({2}\right)}={12}{\left({2}\right)}^{{{3}}}-{25}{\left({2}\right)}^{{{2}}}+{2}+{2}\)

=12(8)-25(4)+4

=96-100+4

=0

So, x=2 is a zero.

Now find the other zeros as

Now

\(\displaystyle{\frac{{{12}{x}^{{{3}}}-{25}{x}^{{{2}}}+{x}+{2}}}{{{x}-{2}}}}={\frac{{{\left({3}{x}-{1}\right)}{\left({4}{x}^{{{2}}}-{7}{x}-{2}\right)}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}^{{{2}}}-{\left({8}-{1}\right)}{x}-{2}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}^{{{2}}}-{8}{x}+{1}{x}-{2}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left[{4}{x}{\left({x}-{2}\right)}+{1}{\left({x}-{2}\right)}\right]}}}{{{x}-{2}}}}\)

\(\displaystyle={\frac{{{\left({3}{x}-{1}\right)}{\left({x}-{2}\right)}{\left({4}{x}+{1}\right)}}}{{{x}-{2}}}}\)

=(3x-1)(4x+1)

Step 3

Hence, the zero are x = 2, \(\displaystyle{\frac{{{1}}}{{{3}}}},{\frac{{-{1}}}{{{4}}}}\)