Factor #9a^5 -4a^3 -81a^2 +36# completely over the a) intergers c) reals b) rationals d) complex numbers ??

2 Answers
Jul 26, 2017

Factor out the first two terms and the next two terms to get an equal factor out.

Explanation:

#9a^5-4a^3-81a^2+36#

Factor the first two an the next two terms out.

#=a^3(9a^2-1)-9(9a^2-4)#

Factor each of the monomials out.

#=(9a^2-1)(a^3-9)#

#=(3a+1)(3a-1)(x-root(3)3)(x^2+root(3)3x+3root(3)3)#

Jul 26, 2017

#9a^5 -4a^3 -81a^2 +36#

#a^3(9a^2-4)-9(9a^2-4)#

#(a^3-9)(9a^2-4)#

#a^3-9# cannot be factored over the integers and #9a^2-4# is a difference of squares, so it can be factored over #ZZ#

#(a^3-9)(3a+2)(3a-2)#

#root(3)9# is irrational and so is #root(3)9^2 = 3root(3)3#

Use the difference of cubes to factor.

#(a-root(3)9)(a^2+aroot(3)9+root(3)9^2)(3a+2)(3a-2)#

The other 2 cube roots of #9# are imaginary and can be found by applying the quadratic formula (of complete the square) for

# a^2+aroot(3)9+root(3)9^2 =0#

#a = (-root(3)9 +- sqrt((root(3)9)^2-4(1)(root(3)9^2)))/(2(1))#

# = (-root(3)9 +- sqrt((3root(3)3)-12root(3)3))/2#

# = (-root(3)9 +- sqrt(-9root(3)3))/2#

# = (-root(3)9 +- 3sqrt(root(3)3)i)/2#

# = (-root(3)9 +- 3root(6)3i)/2#

So we can finish factoring over #CC#

#(a-root(3)9)(a-(-root(3)9 +3root(6)3i)/2)(a-(-root(3)9 - 3root(6)3i)/2)(3a+2)(3a-2)#