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

##### 1 Answer

#### Answer:

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

#= 9r^2+13r+11#

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

#### Explanation:

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

we can calculate the discriminant

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

Since

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

The zeros of

#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 *imaginary unit*, satisfying

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

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