Question

on factor of x^4+x^2+1

Answer 1

x⁴+ 1 + x² 

Add and subtract 2(x²)(1) to make a perfect square 

= (x²)² − 2(x²)(1) + 2(x²)(1) + (1)² + x² 

= (x² + 1)² − 2x² + x² 

= (x² + 1)² − x² 

= (x² + 1)² − (x)² 

Factorize as difference of 2 squares, a² − b² = (a + b)(a − b) 

= [ (x² + 1) + (x) ][ (x² + 1) − (x) ] 

= (x² + 1 + x)(x² + 1 − x) 

= (x² + x + 1)(x² − x + 1) 
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Answer 2

Answer:

Factors of x⁴+x²+1 = (x+1+√x)(x+1-√x)(x²-x+1)

Step-by-step explanation:

x⁴+x²+1

/* Add and subtract x² , we get

= x⁴ +x²+x²-x²+1

= (x²)² + 2x²+1 - x²

= [(x²)² + 2×x²×1 + 1²] - x²

= (x²+1)² - x²

= ( x² + 1 + x )( x²+1-x)

= (x²+2x+1-x)(x²-x+1)

= [(x+1)²-(√x)²](x²-x+1)

= (x+1+√x)(x+1-√x)(x²-x+1)

Therefore,

Factors of x⁴+x²+1 = (x+1+√x)(x+1-√x)(x²-x+1)

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