Find all roots of each of the of the following quartic polynomials:a) z^4+\sqrt{10}z+\frac{1}{4} using factorization into two quadratic polynomials

Kye 2021-09-13 Answered
Find all roots of each of the of the following quartic polynomials:
a) \(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}\) using factorization into two quadratic polynomials

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Expert Answer

Jozlyn
Answered 2021-09-14 Author has 13445 answers
Solution:
Given:
\(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}\)
Let the two quadratic factors be \(\displaystyle{\left({z}^{{2}}+{a}{z}+{b}\right)}\) and \(\displaystyle{\left({z}^{{2}}+{c}{z}+{d}\right)}\)
Then,
\(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}={\left({z}^{{2}}+{a}{z}+{b}\right)}{\left({z}^{{2}}+{c}{z}+{d}\right)}\)
\(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}={z}^{{4}}+{\left({a}+{c}\right)}{z}^{{3}}+{\left({b}+{d}\right)}{z}^{{2}}+{\left({b}{c}+{a}{d}\right)}{z}+{b}{d}\)
Compare the coefficients of power of z on both sides
\(\displaystyle{a}+{c}={0}\)
\(\displaystyle{b}+{d}={0}\)
\(\displaystyle{b}{c}+{a}{d}=\sqrt{{{10}}}\)
\(\displaystyle{b}{d}={\frac{{{1}}}{{{4}}}}\)
Now,
\(\displaystyle{b}{d}={\frac{{{1}}}{{{4}}}}\) gives \(\displaystyle{b}={\frac{{{1}}}{{{4}{d}}}}\), substitute \(\displaystyle{b}={\frac{{{1}}}{{{4}{d}}}}\) into \(\displaystyle{b}+{d}={0}\)
\(\displaystyle{\frac{{{1}}}{{{4}{d}}}}+{d}={0}\)
\(\displaystyle{\frac{{{1}+{4}{d}^{{2}}}}{{{4}{d}}}}={0}\)
which gives
\(\displaystyle{4}{d}^{{2}}=-{1}\)
\(\displaystyle{d}=\pm{\frac{{{i}}}{{{2}}}}\)
Take \(\displaystyle{d}={\frac{{{i}}}{{{2}}}}\), substitute this into \(\displaystyle{b}={\frac{{{1}}}{{{4}{d}}}}\)
\(\displaystyle{b}={\frac{{{1}}}{{{2}{i}}}}\)
\(\displaystyle{b}=-{\frac{{{i}^{{2}}}}{{{2}{i}}}}\)
\(\displaystyle{b}=-{\frac{{{i}}}{{{2}}}}\)
Now, substitute \(\displaystyle{d}={\frac{{{i}}}{{{2}}}}\) and \(\displaystyle{b}=-{\frac{{{i}}}{{{2}}}}\) into \(\displaystyle{b}{c}+{a}{d}=\sqrt{{{10}}}\), we get
\(\displaystyle{\frac{{{i}}}{{{2}}}}{a}-{\frac{{{i}}}{{{2}}}}{c}=\sqrt{{{10}}}\)
Solve: \(\displaystyle{\frac{{{i}}}{{{2}}}}{a}-{\frac{{{i}}}{{{2}}}}{c}=\sqrt{{{10}}}\) and a+c=0
We get \(\displaystyle{a}=-\sqrt{{{10}}}{i}\) and \(\displaystyle{c}=\sqrt{{{10}}}{i}\)
Thus, the quadratic factors are:
\(\displaystyle{\left({z}^{{2}}-\sqrt{{{10}}}{i}{z}-{\frac{{{i}}}{{{2}}}}\right)}\) and \(\displaystyle{\left({z}^{{2}}+\sqrt{{{10}}}{i}{z}+{\frac{{{i}}}{{{2}}}}\right)}\)
Thus,
\(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}={\left({z}^{{2}}-\sqrt{{{10}}}{i}{z}-{\frac{{{i}}}{{{2}}}}\right)}{\left({z}^{{2}}+\sqrt{{{10}}}{i}{z}+{\frac{{{i}}}{{{2}}}}\right)}\)
Find the roots of \(\displaystyle{\left({z}^{{2}}-\sqrt{{{10}}}{i}{z}-{\frac{{{i}}}{{{2}}}}\right)}\):
By the quadratic formula
\(\displaystyle{z}={\frac{{\sqrt{{{10}}}{i}\pm\sqrt{{-{10}-{2}{i}}}}}{{{2}}}}\)
Thus, the roots of \(\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}\) are:
\(\displaystyle{z}={\frac{{\sqrt{{{10}}}{i}+\sqrt{{{10}+{2}{i}}}}}{{{2}}}}\)
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