How do you factor #27-8t^3#?

1 Answer
Jan 30, 2016

Answer:

#=27-8t^3=(-2t+3)(4t^2+6t+9)=(-2t+3)(2t + 3/2 -(3sqrt(3))/2i)(2t+3/2+(3sqrt(3))/2i)#

Explanation:

The difference of cubes formula states that
#a^3 - b^3 = (a-b)(a^2+ab+b^2)#
(try expanding the right hand side to verify this)

Applying the above formula, we have

#27-8t^3 = 3^3-(2t)^3#

#=(3-2t)(3^2+3(2t)+(2t)^2)#

#=(-2t+3)(4t^2+6t+9)#

If we are limiting ourselves to the reals, then we are done, as the discriminant #6^2-4(4)(9)# of #4t^2+6t+9# is less than zero. If we are willing to use complex numbers, then by the quadratic formula, #4t^2+6t+9# has the roots

#t = (-6+-sqrt(-108))/8#

#=(-6+-6sqrt(3)i)/8#

#=-3/4+-(3sqrt(3))/4i#

and so, multiplying by #4# to obtain the correct coefficient for #t#, we have the complete factorization as

#27-8t^3 = (-2t+3)(2t + 3/2 -(3sqrt(3))/2i)(2t+3/2+(3sqrt(3))/2i)#