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

Question

1431 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:

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

How do you factor 6r^3-35r-11r^26r^3-35r-11r^2?