How do you factor #9x^5+6x^4-12x-8#?

1 Answer
Jun 13, 2016

Answer:

#9x^5+6x^4-12x-8=(3x^4-4)(3x+2)#

= #(root(2)3x^2+2)(root(4)3x+sqrt2)(root(4)3x-sqrt2)(3x+2)#

Explanation:

#9x^5+6x^4-12x-8#

= #3x^4(3x+2)-4(3x+2)#

= #(3x^4-4)(3x+2)#

Now no further factors are there if we the domain is rational numbers.

However, if the domain is extended to real i.e. include irrational numbers too, above can be factorized further as below - as #(x^4-a^4)=(x^2+a^2)(x+a)(x-a)#.

#(3x^4-4)(3x+2)#

= #((root(4)3x)^4-(sqrt2)^4)(3x+2)#

= #((root(4)3x)^2+(sqrt2)^2)((root(4)3x)+(sqrt2))((root(4)3x)-(sqrt2))(3x+2)#

= #(root(2)3x^2+2)(root(4)3x+sqrt2)(root(4)3x-sqrt2)(3x+2)#