Question

factorise x^4+1/x^4+1

Answer 1

Answer:

Step-by-step explanation:

x⁴ + 1/x⁴ + 1

=> (x²)² + (1/x²)² + 1

=> (x² + 1/x²)² - 2 + 1

=> (x² + 1/x²)²  - 1

=> (x² + 1/x² - 1) (x² + 1/x² + 1)

=> (x² + 1/x² - 1) ( (x+1/x)² - 1)

=> (x² + 1/x² - 1)( x+1/x + 1)(x + 1/x - 1)

Answer 2

$x^{4}+\frac{1}{x^{4} } +1$ can be factorized as $(x+\frac{1}{x}+1)( x+\frac{1}{x}-1)((x-\frac{1}{x}) ^{2} +1)$.

Step-by-step explanation:

• Here the given polynomial expression is, $x^{4}+\frac{1}{x^{4} } +1$.
• Now, $x^{4}$ and $\frac{1}{x^{4} }$ can also be written as $(x^{2} )^{2}$ and $\frac{1}{(x^{2}) ^{2} }$
• Therefore, the polynomial expression becomes $(x^{2} )^{2}+\frac{1}{(x^{2}) ^{2} } +1$
• Now, we know the well known algebraic identity that is $(a-b)^{2} = a^{2}+b^{2}-2ab\\= > (a-b)^{2}- 2ab= a^{2}+b^{2}$
• Applying this identity in the polynomial given in the problem, we get,

        $(x^{2} +\frac{1}{x^{2} } )^{2} - 2x*\frac{1}{x} +1\\=(x^{2} +\frac{1}{x^{2} } )^{2} -2+1\\=(x^{2} +\frac{1}{x^{2} } )^{2} -1$....(1)

• Now, we know the well known algebraic identity that is

       $a^{2}-b^{2} = (a+b)*(a-b)$

• Applying this formula to the polynomial in (1), we get,

       $(x^{2} +\frac{1}{x^{2} } )^{2} -1 \\=((x^{2} +\frac{1}{x^{2} } )+1)*((x^{2} +\frac{1}{x^{2} } )-1)$

       where a= $x^{2} +\frac{1}{x^{2} }$ and b= 1.

• Again, applying the identity $(a-b)^{2} = a^{2}+b^{2}-2ab\\= > (a-b)^{2}- 2ab= a^{2}+b^{2}$ , we get,

       $((x+\frac{1}{x}) ^{2} -2x\frac{1}{x} +1)((x+\frac{1}{x}) ^{2} -2x\frac{1}{x} -1)\\= ((x+\frac{1}{x}) ^{2} -2+1)((x-\frac{1}{x}) ^{2} +2-1)\\\\=((x+\frac{1}{x}) ^{2} -1)((x-\frac{1}{x}) ^{2} +1)\\\\\\=((x+\frac{1}{x}) ^{2} -1^{2} )((x-\frac{1}{x}) ^{2} +1)\\\\\\\\=(x+\frac{1}{x}+1)( x+\frac{1}{x}-1)((x-\frac{1}{x}) ^{2} +1)$

Hence, $x^{4}+\frac{1}{x^{4} } +1$ can be factorized as $(x+\frac{1}{x}+1)( x+\frac{1}{x}-1)((x-\frac{1}{x}) ^{2} +1)$.

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