Question

How do you factor \[4{x^4} + 81\]?

Answer 1

Hint: We have to find the factor of \[4{x^4} + 81\]. Since, the highest power of \[x\] is four. \[4{x^4} + 81\] will have four roots. To find the roots we will take the square root of \[{x^4}\] to find \[{x^2}\], then the square root of \[{x^2}\] to find the values of \[x\]. We will express the roots in terms of the imaginary unit \[i\] and we will use its properties to simplify the expression. Then at last, we will express all the roots say \[a\], \[b\], \[c\] and \[d\] as \[(x - a)(x - b)(x - c)(x - d)\] which is the factor of \[4{x^4} + 81\].

Complete step by step answer:
Given, \[4{x^4} + 81\].
To find the factor we can write it as
\[ \Rightarrow 4{x^4} + 81 = 0\]
Subtracting \[81\] from both the sides, we get
\[ \Rightarrow 4{x^4} = - 81\]
Dividing both the sides by \[4\], we get
\[ \Rightarrow {x^4} = \dfrac{{ - 81}}{4}\]
Taking square roots of both the sides, we get
\[ \Rightarrow {x^2} = \sqrt {\dfrac{{ - 81}}{4}} \]
On rewriting we get
\[ \Rightarrow {x^2} = \sqrt { - 1} \times \sqrt {\dfrac{{81}}{4}} \]
As we know, \[i = \sqrt { - 1} \]. Using this we get,
\[ \Rightarrow {x^2} = \pm \dfrac{9}{2}i\]
Now as we know,
\[ \Rightarrow \pm \sqrt i = \pm \dfrac{{1 + i}}{{\sqrt 2 }}\]
Multiplying \[\sqrt {\dfrac{9}{2}} \] on both the sides, we get
\[ \Rightarrow \pm \sqrt {\dfrac{9}{2}i} = \pm \sqrt {\dfrac{9}{2}} \dfrac{{1 + i}}{{\sqrt 2 }}\]
On simplifying we get
\[ \Rightarrow \pm \sqrt {\dfrac{9}{2}i} = \pm \dfrac{{3 + 3i}}{2}\]
Similarly,
\[ \Rightarrow \pm \sqrt { - \dfrac{9}{2}i} = \pm \left( i \right) \times \sqrt {\dfrac{9}{2}i} \]
On simplifying we get
\[ \Rightarrow \pm \sqrt { - \dfrac{9}{2}i} = \pm \dfrac{{i\left( {3 + 3i} \right)}}{2}\]
On further simplification and using \[{i^2} = - 1\], we get
\[ \Rightarrow \pm \sqrt { - \dfrac{9}{2}i} = \pm \dfrac{{3i - 3}}{2}\]
Therefore, we can write \[4{x^4} + 81\] as
\[ \Rightarrow 4{x^4} + 81 = 4\left( {x + \dfrac{{3 + 3i}}{2}} \right)\left( {x + \dfrac{{3 - 3i}}{2}} \right)\left( {x - \dfrac{{3 + 3i}}{2}} \right)\left( {x - \dfrac{{3 - 3i}}{2}} \right)\],
\[ \Rightarrow 4{x^4} + 81 = 4\left( {x + \dfrac{3}{2} + \dfrac{{3i}}{2}} \right)\left( {x + \dfrac{3}{2} - \dfrac{{3i}}{2}} \right)\left( {x - \dfrac{3}{2} - \dfrac{{3i}}{2}} \right)\left( {x - \dfrac{3}{2} + \dfrac{{3i}}{2}} \right)\]
Using the identity \[{a^2} - {b^2} = (a - b)(a + b)\], we get
\[ \Rightarrow 4{x^4} + 81 = 2\left( {{{\left( {x + \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{{3i}}{2}} \right)}^2}} \right) \times 2\left( {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{{3i}}{2}} \right)}^2}} \right)\]
On simplification and using \[{i^2} = - 1\], we get
\[ \Rightarrow 4{x^4} + 81 = \left( {2{x^2} - 6x + 9} \right)\left( {2{x^2} + 6x + 9} \right)\]
Therefore, factor of \[4{x^4} + 81\] is \[\left( {2{x^2} - 6x + 9} \right)\left( {2{x^2} + 6x + 9} \right)\].

Note:
We can also solve this problem by another method in which we will use different identities to simplify.
Given, \[4{x^4} + 81\].
We can write \[81\] as \[{3^4}\].
\[ \Rightarrow 4{x^4} + 81 = 4{x^4} + {3^4}\]
On rewriting we get,
\[ \Rightarrow 4{x^4} + 81 = {\left( {2{x^2}} \right)^2} + {\left( 9 \right)^2}\]
As, \[{a^2} + {b^2} = {(a + b)^2} - 2ab\], using this we can write
\[ \Rightarrow 4{x^4} + 81 = {\left( {2{x^2} + 9} \right)^2} - 2 \times 2{x^2} \times 9\]
On simplifying we get,
\[ \Rightarrow 4{x^4} + 81 = {\left( {2{x^2} + 9} \right)^2} - 36{x^2}\]
On rewriting we get,
\[ \Rightarrow 4{x^4} + 81 = {\left( {2{x^2} + 9} \right)^2} - {\left( {6x} \right)^2}\]
Using the identity, \[{a^2} - {b^2} = (a - b)(a + b)\] we get
\[ \Rightarrow 4{x^4} + 81 = \left( {2{x^2} + 9 - 6x} \right)\left( {2{x^2} + 9 + 6x} \right)\]
On rewriting we get,
\[ \Rightarrow 4{x^4} + 81 = \left( {2{x^2} - 6x + 9} \right)\left( {2{x^2} + 6x + 9} \right)\]
Therefore, factor of \[4{x^4} + 81\] is \[\left( {2{x^2} - 6x + 9} \right)\left( {2{x^2} + 6x + 9} \right)\].