# Find all rational zeros of the polynomial, and write the polynomial in factored form. P(x)=12x^{3}-25x^{2}+x+2

Find all rational zeros of the polynomial, and write the polynomial in factored form.
$$\displaystyle{P}{\left({x}\right)}={12}{x}^{{{3}}}-{25}{x}^{{{2}}}+{x}+{2}$$

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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, $$\displaystyle{\frac{{{1}}}{{{3}}}},{\frac{{-{1}}}{{{4}}}}$$