How do you use the Rational Root Theorem to factorise #6x^3+6x^2-72#?

Question

447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-19T04:34:26

#6(x-2)(x^2+3x+6)#

Expression#= 6x^3+6x^2-72#

Clearly #6# is a factor

#:. #Expresion #=6(x^3+x^2-12)#

Applying the rational root theorem to #x^3+x^2-12#

#a_0=12; a_n=1#

The divisors of #a_0# are #1, 2, 3, 4, 6, 12#

The divisor of #a_n# is #1#

So, we must check the following rational numbers as potential roots
of the expression: #+- (1, 2, 3, 4, 6, 12)/1#

We observe that +2 is a root since #2^3+2^2-12 = 8+4-12=0#

Since #+2# is a root #-> (x-2)# is a factor.

Computing by polynomial division #(x^3+x^2-12)/(x-2) -> x^2+3x+6#

Since #x^2+3x+6# does not factorise further

Expression#= 6(x-2)(x^2+3x+6)#