# 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

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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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 $$\displaystyle{c}=\sqrt{{{10}}}{i}$$
$$\displaystyle{\left({z}^{{2}}-\sqrt{{{10}}}{i}{z}-{\frac{{{i}}}{{{2}}}}\right)}$$ 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 $$\displaystyle{\left({z}^{{2}}-\sqrt{{{10}}}{i}{z}-{\frac{{{i}}}{{{2}}}}\right)}$$:
$$\displaystyle{z}={\frac{{\sqrt{{{10}}}{i}\pm\sqrt{{-{10}-{2}{i}}}}}{{{2}}}}$$
Thus, the roots of $$\displaystyle{z}^{{4}}+\sqrt{{{10}}}{z}+{\frac{{{1}}}{{{4}}}}$$ are:
$$\displaystyle{z}={\frac{{\sqrt{{{10}}}{i}+\sqrt{{{10}+{2}{i}}}}}{{{2}}}}$$