Question 156568
<font size = 6 color = "red"><b>EDWIN'S EXPLANATION:</font></b>

How do I solve this (below) using factoring? 
{{{x^2 + 6x - 27 = 0}}} 
I appreciate your help so much. Thank you.
<pre><font size = 4 color = "indigo"><b>
Rule for factoring a trinomial whose coefficient
of the squared letter is understood as +1.

From right to left look at these:

P. The last number in absolute value.
Q. The sign before the last number.
R. The middle coefficient in absolute value.
S. The sign before the middle coefficient.

1. Think of a pair of factors of P, such that when you 
do the operation determined by Q to them, you get R. 
Then write

(x    )(x    )

2. Place the two factors where the #'s are below:

(x   #)(x   #)

3. Place the sign from S before the larger of the #'s

4. If Q is +, place the same sign as S on the smaller of the #'s.

5. If Q is -, place the opposite sign from S in the smaller of the #'s. 

---------

Let's go through your problem with these to factor
the left side:

{{{x^2 + 6x - 27 = 0}}}

From right to left look at these:

P. The last number in absolute value.  THIS IS 27
Q. The sign before the last number.  THIS IS -
R. The middle coefficient in absolute value.  THIS IS 6
S. The sign before the middle coefficient.  THIS IS +

1. Think of a pair of factors of P, such that when you 
do the operation determined by Q to them, you get R. 

THESE ARE 9 AND 3, because 9x3=27 and 9-3=6
 
Then write

(x    )(x    )

2. Place the two factors where the #'s are below:

(x   #)(x   #)

SO WE WRITE (x   9)(x   3)


3. Place the sign from S before the larger of the #'s

THE LARGER IS 9, AND THE SIGN FROM S is +, 

SO WE WRITE: (x + 9)(x   3)

 
4. If Q is +, place the same sign on the smaller of the #'s.

IT'S NOT.


5. If Q is -, place the opposite sign in the smaller of the #'s. 

SO WE WRITE (x + 9)(x - 3)

Now we have gone from this:

{{{x^2 + 6x - 27 = 0}}}

to this

{{{(x + 9)(x - 3)=0}}}

Now to use the zero-factor principle:

Each of the parentheses on the left
represents a number.  They are multipled
to get 0 on the right. In order to get a 
0 on the right one of those two parentheses
must equal 0. It could be either one, so
we get two solutions, by setting each
factor on the left = 0 and solving for x

{{{x+9=0}}} gives the solution {{{x=-9}}}

{{{x-3=0}}} gives the solution {{{x=3}}}

Edwin</pre>