Question

find the hcf of x²_x+1;x⁴+x​

Answer 1

Answer:

x³ – 1= x³ - 1³

= (x - 1)(x²+x+1)

x² + x² + 1 = (x² + 2x² + 1) - x²

= (x² + 1)² - (x)²2

= (x² +1+x)(x² + 1 - x)

= (x² + x + 1)(x² − x + 1)

Hence, H.C.F. of x³ - 1 and x¹ + x² + 1 is

x² + x + 1.