How do you factor the expression #4x^4-6x^2+2#?

1 Answer
May 12, 2018

Answer:

#4x^4-6x^2+2 = (x-1)(x+1)(2x-sqrt(2))(2x+sqrt(2))#

#color(white)(4x^4-6x^2+2) = 4(x-1)(x+1)(x-sqrt(2)/2)(x+sqrt(2)/2)#

Explanation:

Given:

#4x^4-6x^2+2#

Note that the sum of the coefficients is zero, i.e. #4-6+2=0#.

So we can tell that #x=+-1# are zeros and #(x-1)(x+1) = x^2-1# is a factor.

Also all of the terms are divisible by #2#, so we could separate that out as a factor, but given what we find, it's probably better to leave it in there until the end:

#4x^4-6x^2+2 = (x^2-1)(4x^2-2)#

#color(white)(4x^4-6x^2+2) = (x^2-1^2)((2x)^2-(sqrt(2))^2)#

#color(white)(4x^4-6x^2+2) = (x-1)(x+1)(2x-sqrt(2))(2x+sqrt(2))#

#color(white)(4x^4-6x^2+2) = 4(x-1)(x+1)(x-sqrt(2)/2)(x+sqrt(2)/2)#