Solve by factoring: x^4-12x^2 = 64

The final solution is all the values that make (x+4) (x - 4) (x2 + 4) = 0 x+4 ⁢ x - 4 ⁢ x2 + 4 = 0 true. x = - 4,4, 2i, - 2 i

Minus 64 from both sides

x⁴-12x²-64=0

hmm

what 2 numbers multiply to - 64 and add to get - 12

-16 and 4

(x²-16) (x²+4) = 0

oh look a difference of 2 perfect squares

(x²-4²) (x²+4) = 0

(x-4) (x+4) (x²+4) = 0

set each to zero

x-4=0

x=4

x+4=0

x=-4

x²+4=0

x²=-4

if you have learend complex roots then

sqrt both sides to get

x=-2i or 2i

solutions are

x=-4, 4, 2i or - 2i