Question

Can you show that ${{z}^{4}}+64$ can be factorized into the two real quadratic factors of the form ${{z}^{2}}+az+8$ and ${{z}^{2}}+bz+8$ but cannot be factored into two real quadratic factors of the form ${{z}^{2}}+bz+16$ and ${{z}^{2}}+bz+4$?

Answer 1

Hint: we have to show the real quadratic factors of ${{z} ^ {4}} +64$. Before solving the question we should know about the $z$. Here the given $z$ is the complex number and $4$ is the power of that complex number. Factors of complex numbers mean every polynomial can be factored into a product of linear factors or we can say every polynomial of degree n has n roots, counted according to their multiplicity.

Complete step by step solution:
Now the given question is:
$\Rightarrow {{z} ^ {4}} +64$
The factors of the given equation is
$\Rightarrow {{z}^{4}}+64=\left( z-{{z}_{0}} \right)\left( z-{{z}_{1}} \right)\left( z-{{z}_{2}} \right)\left( z-{{z}_{3}} \right)$
Now puts the above equation equals to zero
$\Rightarrow {{z}^{4}}+64=\left( z-{{z}_{0}} \right)\left( z-{{z}_{1}} \right)\left( z-{{z}_{2}} \right)\left( z-{{z}_{3}} \right)=0$
On solving the above part, we get
$\Rightarrow {{z} ^ {4}} =-64$
Here we can write $64$ in the power of $2$, we get
$\Rightarrow {{z}^{4}}=-64={{2}^{6}}{{e}^{\iota \pi +\iota 2k\pi }}$
Here by using De Moivre's identity we can write
$\Rightarrow {{e} ^ {\iota \pi}}=\cos \pi +\iota \sin \pi =-1$
Now again solving we get
$z=\sqrt[4]{{{2}^{6}}}{{e}^{\iota \left( \dfrac{\pi }{4}+k\dfrac{\pi }{2} \right)}}$
Now we will obtain these for $k=0, 1, 2, 3$
Therefore by using the factors of $z$ we get,
$\Rightarrow {{z}_{0}}=\sqrt[4]{{{2}^{6}}}{{e}^{\iota \dfrac{\pi }{2}}}=\sqrt[4]{{{2}^{6}}}\left( \cos \dfrac{\pi }{4}+\iota \sin \dfrac{\pi }{4} \right)=\sqrt[4]{{{2}^{6}}}\dfrac{1+\iota }{\sqrt{2}}$
$\Rightarrow {{z}_{1}}=\sqrt[4]{{{2}^{6}}}{{e}^{\iota \left( \dfrac{\pi }{4}+\dfrac{\pi }{2} \right)}}$
Now apply the De Moivre`s law, we get
$\begin {align}
  & \Rightarrow {{z}_{1}}=\sqrt[4]{{{2}^{6}}}{{e}^{\iota \dfrac{3\pi }{4}}} \\
 & \Rightarrow {{z}_{1}}=\sqrt[4]{{{2}^{6}}}\left( \cos \dfrac{3\pi }{4}+\iota \sin \dfrac{3\pi }{4} \right) \\
 & \Rightarrow {{z}_{1}}=-\sqrt[4]{{{2}^{^{6}}}}\dfrac{1+\iota }{\sqrt{2}} \\
\end{align}$
Similarly we can find,
$\Rightarrow {{z}_{2}}=-\sqrt[4]{{{2}^{6}}}\dfrac{1+\iota }{\sqrt{2}}$ and$\Rightarrow {{z}_{3}}=\sqrt[4]{{{2}^{6}}}\dfrac{1-\iota }{\sqrt{2}}$
Now arranging the factors
$\begin {align}
  & \Rightarrow \left (z-{{z} _ {0}} \right)\left (z-{{z} _ {3}} \right)={{z}^{2}}-4z+8 \\
 & \Rightarrow \left (z-{{z} _ {1}} \right)\left (z-{{z} _ {2}} \right)={{z}^{2}}+4z+8 \\
\end{align}$
Hence the above are the only to give trinomials with real coefficients. The solution is:
$\Rightarrow {{z}^{4}}+64=\left( {{z}^{2}}-4z+8 \right)\left( {{z}^{2}}+4z+8 \right)$

Note:
We can also find the real factors of the given equation ${{z} ^ {4}} +64$ by grouping coefficients.
Write the above given equation and it's real factors as:
$\begin {align}
  & \Rightarrow {{z}^{4}}+64=\left( {{z}^{2}}-az+8 \right)\left( {{z}^{2}}+bz+8 \right) \\
 & \Rightarrow {{z}^{4}}+64-\left( {{z}^{2}}-az+8 \right)\left( {{z}^{2}}+bz+8 \right)=0 \\
\end{align}$
On solving the above equations we get two equations
$\begin {align}
  & \Rightarrow 16+ab=0 \\
 & \Rightarrow a+b=0 \\
\end{align}$
On solving above these we get
$a=4, b=-4$
Hence we get the two real quadratic factors of given equation is
$\Rightarrow {{z}^{4}}+64=\left( {{z}^{2}}-4z+8 \right)\left( {{z}^{2}}+4z+8 \right)$
Now again if we apply the grouping coefficient rule
$\Rightarrow {{z}^{4}}+64-\left( {{z}^{2}}+bz+16 \right)\left( {{z}^{2}}+bz+4 \right)=0$ we get
$\Rightarrow {{z}^{4}}+64=\left( {{z}^{2}}+bz+16 \right)\left( {{z}^{2}}+bz+4 \right)$
On solving the above equation we get these two equations:
$\begin {align}
  & \Rightarrow b=0, \\
 & \Rightarrow 20+ {{b} ^ {2}} =0 \\
\end{align}$
Therefore, by solving these two we do not get any solution.
Hence ${{z}^{4}}+64$ can be factored into two real quadratic factors $\left( {{z}^{2}}-4z+8 \right)\left( {{z}^{2}}+4z+8 \right)$.