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
ANSWERED
asked 2021-09-19
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

Expert Answers (1)

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
9
 
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