Question

How do you factor completely \[{{m}^{8}}+\dfrac{{{m}^{4}}}{16}+\dfrac{1}{256}\]?

Answer 1

Hint: To factorize this expression, we should know the following expansion property, which states that \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-kab \right)\left( {{a}^{2}}+{{b}^{2}}+kab \right)\]. We will use this expansion to factor the given expression.

Complete step by step answer:
We are given the expression, \[{{m}^{8}}+\dfrac{{{m}^{4}}}{16}+\dfrac{1}{256}\].
Let \[n=\dfrac{1}{2}\], squaring both sides of the above expression, we get \[{{n}^{2}}=\dfrac{1}{4}\]. Squaring once again we get, \[{{n}^{4}}=\dfrac{1}{16}\]. And, squaring once again we get \[{{n}^{8}}=\dfrac{1}{256}\]. We will use these assumptions later.
The given expression \[{{m}^{8}}+\dfrac{{{m}^{4}}}{16}+\dfrac{1}{256}\] can also be expressed as, \[{{m}^{8}}+{{m}^{4}}\times \dfrac{1}{16}+\dfrac{1}{256}\].
Substituting \[{{n}^{4}}\] for \[\dfrac{1}{16}\], and \[{{n}^{8}}\] for \[\dfrac{1}{256}\]. The above expression becomes
\[\Rightarrow {{m}^{8}}+{{m}^{4}}{{n}^{4}}+{{n}^{8}}\]
\[\Rightarrow {{\left( {{m}^{2}} \right)}^{4}}+{{\left( {{m}^{2}} \right)}^{2}}{{\left( {{n}^{2}} \right)}^{2}}+{{\left( {{n}^{2}} \right)}^{4}}\]
The above expression is of the form \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}\]. Comparing the coefficients to find the value of \[k\], we get
\[\Rightarrow 2-{{k}^{2}}=1\]
Subtracting 2 from both sides of above equation, we get
\[\begin{align}
  & \Rightarrow 2-{{k}^{2}}-2=1-2 \\
 & \Rightarrow -{{k}^{2}}=-1 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{k}^{2}}=1 \\
 & \therefore k=\pm 1 \\
\end{align}\]
It doesn’t really matter whether we take positive value or negative value, because of the expansion expression. So, we take \[k=1\].

We know that the expansion of \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}\] is \[\left( {{a}^{2}}+{{b}^{2}}-kab \right)\left( {{a}^{2}}+{{b}^{2}}+kab \right)\].
Using this expansion for the expression \[{{\left( {{m}^{2}} \right)}^{4}}+{{\left( {{m}^{2}} \right)}^{2}}{{\left( {{n}^{2}} \right)}^{2}}+{{\left( {{n}^{2}} \right)}^{4}}\], we get \[\left( {{\left( {{m}^{2}} \right)}^{2}}+{{\left( {{n}^{2}} \right)}^{2}}-1\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\left( {{\left( {{m}^{2}} \right)}^{2}}+{{\left( {{n}^{2}} \right)}^{2}}+1\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\]
\[\Rightarrow \left( {{m}^{4}}+{{n}^{4}}-\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\left( {{m}^{4}}+{{n}^{4}}+\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\]
We have factorized the expression once; we get two factors as \[\left( {{m}^{4}}+{{n}^{4}}-\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\And \left( {{m}^{4}}+{{n}^{4}}+\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\]. By observing these two factors, we notice that both of these factors are of the form \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}\]. Hence, both of them can be further factorized.
For \[\left( {{m}^{4}}+{{n}^{4}}-\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\], comparing with \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}\], we get
\[\Rightarrow 2-{{k}^{2}}=-1\]
Subtracting 2 from both sides, we get
\[\begin{align}
  & \Rightarrow 2-{{k}^{2}}-2=-1-2 \\
 & \Rightarrow -{{k}^{2}}=-3 \\
 & \Rightarrow {{k}^{2}}=3 \\
\end{align}\]
We will take only positive values like before, taking square root of both side we get
\[\therefore k=\sqrt{3}\]
Expansion of \[\left( {{m}^{4}}+{{n}^{4}}-\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\] is \[\left( {{m}^{2}}+{{n}^{2}}-\sqrt{3}mn \right)\left( {{m}^{2}}+{{n}^{2}}+\sqrt{3}mn \right)\]
For \[\left( {{m}^{4}}+{{n}^{4}}+\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\], comparing with \[{{a}^{4}}+{{b}^{4}}+(2-{{k}^{2}}){{a}^{2}}{{b}^{2}}\], we get
\[\Rightarrow 2-{{k}^{2}}=1\]
Subtracting 2 from both sides we get,
\[\begin{align}
  & \Rightarrow 2-{{k}^{2}}-2=1-2 \\
 & \Rightarrow -{{k}^{2}}=-1 \\
 & \Rightarrow {{k}^{2}}=1 \\
 & \therefore k=1 \\
\end{align}\]
Hence, expansion of \[\left( {{m}^{4}}+{{n}^{4}}+\left( {{m}^{2}} \right)\left( {{n}^{2}} \right) \right)\] is \[\left( {{m}^{2}}+{{n}^{2}}-mn \right)\left( {{m}^{2}}+{{n}^{2}}+mn \right)\].
Hence the complete factorization of \[{{m}^{8}}+{{m}^{4}}{{n}^{4}}+{{n}^{8}}\] is \[\left( {{m}^{2}}+{{n}^{2}}-\sqrt{3}mn \right)\left( {{m}^{2}}+{{n}^{2}}+\sqrt{3}mn \right)\left( {{m}^{2}}+{{n}^{2}}-mn \right)\left( {{m}^{2}}+{{n}^{2}}+mn \right)\]
Here \[n=\dfrac{1}{2}\], substituting it can make the equation look too complex. So, we’ll leave it like this.

Note:
 It should be noted that we factorized the given expression twice. We did it because the question says to completely factorize the expression. It means that, to factorize till it cannot be further factorized. If we leave the expression, after factoring once it is not correct, and our answer becomes wrong. The expansion we used to solve is important, and it should be remembered. Calculation mistakes should be avoided.