How do you factor #y= 16x^4-81# ?

1 Answer
Mar 18, 2016

Answer:

Use the difference of squares identity to find:

#y = 16x^4-81#

#=(2x-3)(2x+3)(4x^2+9)#

#=(2x-3)(2x+3)(2x-3i)(2x+3i)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this twice as follows:

#y = 16x^4-81#

#=(4x^2)^2-9^2#

#=(4x^2-9)(4x^2+9)#

#=((2x)^2-3^2)(4x^2+9)#

#=(2x-3)(2x+3)(4x^2+9)#

The remaining quadratic factor has no linear factors with Real coefficients, but if you allow Complex coefficients you can find:

#=(2x-3)(2x+3)((2x)^2-(3i)^2)#

#=(2x-3)(2x+3)(2x-3i)(2x+3i)#