How do you factor #6r^3-35r-11r^2#?

Question

1080 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-16T20:32:46

First, we must recognize all the common elements in your three parts, which, in this case, is just #r#, as follows:

#6*color(green)(r)*r*r*-35*color(green)(r)-11*color(green)(r)*r#

So, we take it out an mutiply the rest by this common factor:

#r(6r^2-35-11r)#

Now, solving this quadratic...

#(-b+-sqrt(b^2-4ac))/2a#
#(11+-sqrt(121-4*6*(-35)))/(2*6)#
#(11+-sqrt(961))/12#
#(11+-31)/12#

#x_1=7/2#, which is the same as #2x_1-7=0#.
#x_2=-5/3#, which is the same as #3x_2+5=0#.

So, now we have our factors, using our roots:

#r(2x-7)(3x+5)#