1: ![3n^{2}+9n+6](https://tex.z-dn.net/?f=3n%5E%7B2%7D%2B9n%2B6)
notice that each part is divisible by 3
÷ 3 = ![n^{2}](https://tex.z-dn.net/?f=n%5E%7B2%7D)
9n ÷ 3 = 3n
6 ÷ 3 = 2
so it becomes ![3(n^{2} +3n+2)](https://tex.z-dn.net/?f=3%28n%5E%7B2%7D%20%2B3n%2B2%29)
3n can be rewritten as 2n+n
-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)
which would get you
![3(n^{2} +2n+n+2)](https://tex.z-dn.net/?f=3%28n%5E%7B2%7D%20%2B2n%2Bn%2B2%29)
Now that it's rewritten, you can factor out n + 2 from the equation.
<u><em>the answer is </em></u>
3(n+2)(n+1)
And you can check that by multiplying (n+2)(n+1) which is
and then each of those by 3, which is
or
, our origional equation
2: ![28+x^{2} -11x](https://tex.z-dn.net/?f=28%2Bx%5E%7B2%7D%20-11x)
So I rewrote this as
(it's the same thing, just reordered using the commutative property)
now -11x can be rewritten as -4x-7x
(remember, the two numbers should multiply to equal 28, which is our constant.)
![x^{2} -4x-7x+28](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-4x-7x%2B28)
now we can factor out x from the first expression and -7 from the second
![x(x-4)-7(x-4)](https://tex.z-dn.net/?f=x%28x-4%29-7%28x-4%29)
and lastly you factor out x-4,
<u><em>which would give you</em></u>
(x-4)(x-7)
Make sure to check your work and make sure it multiplies to ![x^{2} -11x+28](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-11x%2B28)
3: ![9x^{2} -12x+4](https://tex.z-dn.net/?f=9x%5E%7B2%7D%20-12x%2B4)
The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12
So if you rewrite 9 as
and 4 as
, the equation becomes
![3^{2} x^{2} -12x+2^{2}](https://tex.z-dn.net/?f=3%5E%7B2%7D%20x%5E%7B2%7D%20-12x%2B2%5E%7B2%7D)
now that
is ugly so it can be turned into ![(3x)^{2}](https://tex.z-dn.net/?f=%283x%29%5E%7B2%7D)
and -12x can be rewritten as ![-2*3x*2](https://tex.z-dn.net/?f=-2%2A3x%2A2)
so our equation now looks like ![(3x)^2-2*3x*2+2^{2}](https://tex.z-dn.net/?f=%283x%29%5E2-2%2A3x%2A2%2B2%5E%7B2%7D)
There's a rule that says
In our case, a=3x and b=2
<u><em>so the final answer is</em></u>
![(3x-2)^2](https://tex.z-dn.net/?f=%283x-2%29%5E2)