How do you factor #y= 16x^4-81# ?
1 Answer
Mar 18, 2016
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)#
