Question

factorise c^4 - 23c^2 + 1

Answer 1

c⁴ - 23c² + 1

= (c²)² + 2c² + 1 - 25c²

= ( c² + 1)² - (5c)²

use , a² - b² = (a - b)( a + b)

= { c² + 1 - 5c)( c² + 1 + 5c)

now ,
we have two factors ,
( c² + 1 - 5c) and ( c² + 1 + 5c)
both are quadratic , we are also seeing that can't be factorise by simple factorization method now ,
we use ,
quadratic formula for factorisation ,

for , ( c² + 1 - 5c)

c = { 5 ±√ (25 -4 ) }/2
c = { 5 ±√21 }/2

hence,
factor of ( c² - 5c +1) are { c - (5+√21)/2} and { c - ( 5 -√21)/2}

similarly use formula for (c² + 5c +1)
and you will get factors of ( c² + 5c +1) are { c - ( -5+√21)/2 } and { c - ( -5 -√21)/2}


hence, factors of ( c⁴ -23c² + 1) are
{ c - ( 5 +√21)/2 } , { c -(5 -√21)/2 }, { c -( -5+√21)/2} and { c - ( -5-√21)/2}

Answer 2

c⁴ - 23c² + 1
= (c²)² + 2c² + 1 - 25c²
= ( c² + 1)² - (5c)²
use , a² - b² = (a - b)( a + b) = { c² + 1 - 5c)( c² + 1 + 5c)