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

2 Answers
Peter L.
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)

Jim H
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)

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