Question

How do you factor \[{{x}^{4}}+64\] ?

Answer 1

Hint: We have a biquadratic equation with the highest degree coefficient unity and all other variable coefficients are zero. First, equate this coefficient with zero then just do some algebraic operations and note that the square root of minus one is iota \[(i)\] then just simplify the radical root of the number \[64\].

Complete solution:
Since it is given that a polynomial \[{{x}^{4}}+64\], we have to factorize this
First, equate this polynomial with zero and make it an equation
\[\Rightarrow {{x}^{4}}+64=0\]
\[\Rightarrow {{x}^{4}}=-64\]
Now this is of some square type
\[\Rightarrow \] Let \[{{x}^{2}}=u\Rightarrow {{x}^{4}}={{u}^{2}}\]
\[\Rightarrow {{u}^{2}}=-64\]
Now, this negative number can be written as the product of two numbers one of them is positive as the main number is negative.
\[\Rightarrow {{u}^{2}}=-1\times 64\]
\[\Rightarrow u=\sqrt{-1\times 64}\]
And we know that, \[\sqrt[n]{ab}=\sqrt[n]{a}.\sqrt[n]{b}\]
\[\Rightarrow u=\sqrt{-1}\times \sqrt{64}\]
And we know that \[\sqrt{-1}\] is known as the complex root and that is
\[\Rightarrow \sqrt{-1}=i\] , where \[i\] is \[iota\] called complex root
\[\Rightarrow u=i\times \sqrt{64}\]
Now we have to calculate ethe square root of \[64\]
\[64=8\times 8\]
\[\Rightarrow \sqrt{64}=8\]
Sine we assumed a new variable \[u\] as \[{{x}^{2}}\]
\[\Rightarrow u=8i\text{ },\text{ -8i}\]
So, substitute back \[u\] as \[{{x}^{2}}\]
\[\Rightarrow {{x}^{2}}=8i\text{ , -8}i\]
Now continuing with \[8i\]
\[{{x}^{2}}=8i\]
On further simplifying, we get
\[x=2+2i\] and \[x=-2-2i\]
And when we continuing with \[-8i\]
\[x=-2+2i\] and \[x=2-2i\]
Thus, on simplifying \[{{x}^{4}}+64=0\] we get
 \[x=2+2i\] , \[x=-2-2i\], \[x=-2+2i\] and \[x=2-2i\]
Since these are the roots means these values for \[x\] equation will satisfy.
Now for factors,
\[(x-2-2i)\text{ , (x+2+2i) , (x+2-2i) , (x-2+2i)}\]

Hence, \[{{x}^{4}}+64=(x-2-2i)\text{(x+2+2i)(x+2-2i)(x-2+2i)}\]

Note:
When we are given an equation with discriminant zero that means real roots don’t exist but complex roots exist. So, using the property of a complex root that is the square root of minus one is iota. And also remember that the property \[\sqrt[n]{ab}=\sqrt[n]{a}.\sqrt[n]{b}\] is only applicable when \[a\] and \[b\] both are of opposite signs.