How do you factor y= 16x^4-81y=16x4−81 ?
1 Answer
Mar 18, 2016
Use the difference of squares identity to find:
y = 16x^4-81y=16x4−81
=(2x-3)(2x+3)(4x^2+9)=(2x−3)(2x+3)(4x2+9)
=(2x-3)(2x+3)(2x-3i)(2x+3i)=(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)a2−b2=(a−b)(a+b)
Use this twice as follows:
y = 16x^4-81y=16x4−81
=(4x^2)^2-9^2=(4x2)2−92
=(4x^2-9)(4x^2+9)=(4x2−9)(4x2+9)
=((2x)^2-3^2)(4x^2+9)=((2x)2−32)(4x2+9)
=(2x-3)(2x+3)(4x^2+9)=(2x−3)(2x+3)(4x2+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)2−(3i)2)
=(2x-3)(2x+3)(2x-3i)(2x+3i)=(2x−3)(2x+3)(2x−3i)(2x+3i)
