How do you factor 6z^3-3z^2-30z6z3−3z2−30z?

1 Answer
Lovecraft
Oct 3, 2015

3z(2z-5)(z+2)3z(2z−5)(z+2)

Explanation:

Factor the the GCF, which is 3z3z

3z(2z^2 -1z -10)3z(2z2−1z−10)

To see if the quadratic can be factored any further, find the two roots r_1r1 and r_2r2, if they exist the quadratic will be able to be written in the form of (z-r_1)(z-r_2)(z−r1)(z−r2).

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

z = (1+- sqrt(1-(-80)))/4 = (1+- sqrt(81))/4z=1±√1−(−80)4=1±√814
z = (1 +- 9)/4z=1±94

So we have
r_1 = (1+9)/4 = 10/4 = 5/2r1=1+94=104=52
r_2 = (1-9)/4 = -8/4 = -2r2=1−94=−84=−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)3z(2z−5)(z+2)

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