Question

# Find the zeros ofcthe polynomials f(x)=x^3-12x^2+39x-28, if it's is given that the zeros in AP

Polynomials
Find the zeros ofcthe polynomials $$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}-{12}{x}^{{2}}+{39}{x}-{28}$$, if it's is given that the zeros in AP

2021-09-20
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}-{12}{x}^{{2}}+{39}{x}-{28}$$
Let x,y,z are zeros of given polynomial.
we know that,
if x,y,z, are zero of cubic polynomial $$\displaystyle{a}{x}^{{3}}+{b}{x}^{{2}}+{c}{x}+{d}$$
then,
$$\displaystyle{x}+{y}+{z}={\frac{{-{b}}}{{{a}}}}$$
$$\displaystyle{x}{y}+{y}{z}+{z}{x}={\frac{{{c}}}{{{a}}}}$$
$$\displaystyle{x}{y}{z}={\frac{{-{d}}}{{{a}}}}$$
now given polynomial is,
$$\displaystyle{x}^{{3}}-{12}{x}^{{2}}+{39}{x}-{28}$$
Here, $$\displaystyle{a}={1},{b}=-{12},{c}={39},{d}=-{28}$$
$$\displaystyle{x}+{y}+{z}={\frac{{-{b}}}{{{a}}}}={\frac{{-{12}}}{{{1}}}}={12}$$
$$\displaystyle{x}{y}+{y}{z}+{z}{x}={\frac{{{c}}}{{{a}}}}={\frac{{{39}}}{{{1}}}}={39}$$
$$\displaystyle{x}{y}{z}=-{\frac{{{d}}}{{{a}}}}=-{\frac{{-{28}}}{{{1}}}}={28}$$
Now, given that polynomial zeros are in A.P.,
$$\displaystyle{x}={a}-{d}.{y}={a}.{z}={a}+{d}$$
put these value of x,y,z in equation, we get
$$\displaystyle{\left({a}-{d}\right)}+{a}+{\left({a}+{d}\right)}={12}\Rightarrow{3}{a}={12}\Rightarrow{a}={4}$$
put value of x,y,z in equation we get
$$\displaystyle{\left({a}-{d}\right)}+{a}+{\left({a}+{d}\right)}={28}\Rightarrow{a}{\left({a}^{{2}}-{d}^{{2}}\right)}={28}$$
$$\displaystyle\Rightarrow{a}{\left({a}^{{2}}-{d}^{{2}}\right)}={28}$$
now a=4, $$\displaystyle{4}{\left({16}-{d}^{{2}}\right)}={28}$$
$$\displaystyle{d}^{{2}}={9}\Rightarrow{d}=\pm{3}$$
we know, $$\displaystyle{x}={a}-{d}={4}-{5}={1}$$
$$\displaystyle{y}={a}={4}$$
$$\displaystyle{z}={a}+{d}={4}+{3}={7}$$
The polynomial zero are, 1,4,7