How do you factor #6r + 8- r + 8r + 9r ^ { 2} + 3#?

Given:

#6r+8-r+8r+9r^2+3#

First rearrange into standard form, in decreasing order of degree, combining terms of the same degree:

#6r+8-r+8r+9r^2+3 = 9r^2+6r-r+8r+8+3#

#color(white)(6r+8-r+8r+9r^2+3) = 9r^2+13r+11#

Now this is in standard form:

#ar^2+br+c#

with #a=9#, #b=13# and #c=11#

we can calculate the discriminant #Delta# using the formula:

#Delta = b^2-4ac = 13^2-4(9)(11) = 169 - 396 = -227#

Since #Delta < 0#, this quadratic has no linear factors with real coefficients.

If you want to factor it with Complex coefficients then you can do so with the help of the quadratic formula.

The zeros of #9r^2+13r+11# are given by the formula:

#r = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(r) = (-b+-sqrt(Delta))/(2a)#

#color(white)(r) = (-13+-sqrt(-227))/18#

#color(white)(r) = -13/18+-sqrt(227)/18i#

(where #i# is the imaginary unit, satisfying #i^2=-1#)

Hence, two factors of our quadratic can be written:

#(r+13/18+sqrt(227)/18i)#

and

#(r+13/18-sqrt(227)/18i)#

In order for the coefficient of the leading term to be #9#, we need to multiply the product of these by #9# to find:

#9r^2+13r+11 = 9(r+13/18+sqrt(227)/18i)(r+13/18-sqrt(227)/18i)#