How do you factor #6z^3-3z^2-30z#?

1 Answer
Oct 3, 2015

Answer:

#3z(2z-5)(z+2)#

Explanation:

Factor the the GCF, which is #3z#

#3z(2z^2 -1z -10)#

To see if the quadratic can be factored any further, find the two roots #r_1# and #r_2#, if they exist the quadratic will be able to be written in the form of #(z-r_1)(z-r_2)#.

The roots are #z = (1+- sqrt(1-4*2*(-10)))/4#, from that we get

#z = (1+- sqrt(1-(-80)))/4 = (1+- sqrt(81))/4#
#z = (1 +- 9)/4#

So we have
#r_1 = (1+9)/4 = 10/4 = 5/2#
#r_2 = (1-9)/4 = -8/4 = -2#

Now, since the parabola there has a two as the leading coefficient and we have a root who's a fraction two, we can say that the factored form is

#3z(2z-5)(z+2)#