$\displaystyle x^2-12x+36$ fits a common model of a special class of polynomial because the last number, $\displaystyle 36$, is half the middle term,$\displaystyle -6$ squared (i.e. take $\displaystyle 1/2$ the middle number and square it and you have the last term in the polinomial.) This is very similar to the process of "completing the square" of a quadratic.

$\displaystyle x^2-12x+36$

$\displaystyle =x^2+2*(\underline{-6})x+(\underline{-6})^2$

$\displaystyle =(x+(\underline{-6}))(x+(\underline{-6}))$

$\displaystyle =(x-6)(x-6)$

$\displaystyle =(x-6)^2$

The goal is to factor out the greatest common factor to leave the polynomial in a much cleaner state. We notice that there is a factor of $\displaystyle 2$ and $\displaystyle b$ (it could be that $\displaystyle b$ is a number or a variable but, in this case, it doesn't matter). We can pull out $\displaystyle 2b$ from all of the terms and put it in front.

$\displaystyle 2bx-4by+6bz$

$\displaystyle =\underline{2b}x-2*\underline{2b}y+3*\underline{2b}z$

Factoring out the 2b and leaving what's left inside of the parentheses, we get:

$\displaystyle =2b(x-2y+3z)$

Note that this can't be simplified or factored anymore because there are no more common factors within the parentheses.

We notice that this is the difference of two squared numbers: $\displaystyle x^2$ and $\displaystyle 8^2$.

Hence, we can follow the rule that the difference of two perfect squares $\displaystyle 8^2-x^2$ is equal to$\displaystyle (8+x)(8-x)$.

To see this a little better, we can FOIL out the answer:

$\displaystyle (8+x)(8-x)$

$\displaystyle =(8^2)-8x+8x+(x)*(-x)$

$\displaystyle =64-8x+8x-(x^2)$

the $\displaystyle 8x$s cancel out and we're left with the original equation:

$\displaystyle 64-x^2$

Remember that whenever there's a problem involving factoring, you can always expand your answer again and see if you end up with the original expression given.

The first thing to notice is that the polynomial $\displaystyle 9x^2-180x+900$ has a common factor of $\displaystyle 9$ so we can factor it out automatically.

$\displaystyle 9x^2-180x+900$

$\displaystyle =9(x^2-20x+100)$

From here, we have a reducible quadratic factor in the parentheses. We know this because we consider the middle term: Half of the middle term squared is equal to the last term. Let's see this together: half of  the middle term, $\displaystyle 20$, is $\displaystyle 10$. $\displaystyle 10^2$ is $\displaystyle 100$ and equal to the last term.

 That means that we can factor the polynomial thusly:

$\displaystyle =9(x^2+2*(\underline{-10})x+(\underline{10})^2)$

$\displaystyle =9(x+(-10))^2$

$\displaystyle =9(x-10)^2$ 

$\displaystyle checking:$

To check to see if our answer is correct, we can expand it again to see if we end up with the original polynomial.

$\displaystyle 9(x-10)^2$

$\displaystyle =9(x-10)(x-10)$

Expanding the two linear factors using FOIL

$\displaystyle =9(x^2-10x-10x+(-10)^2$

$\displaystyle =9(x^2-20x+100)$

Distributing out the 9 in front, we have the original polynomial.

$\displaystyle =9x^2-180x+900$

This one's tricky. We must pull out the greatest common factor from the polynomial first to see what we end up with. It looks like each of the terms has a factor of $\displaystyle 4$, $\displaystyle a$, and $\displaystyle x$. That means we can pull out $\displaystyle 4ax$ from each factor and put it in front of the parentheses.

$\displaystyle 4ax^3-48ax^2+144ax$

$\displaystyle =(\underline{4ax})*x^2-(\underline{4ax})*12x+(\underline{4ax})*36$

$\displaystyle =4ax(x^2-12x+36)$

Now, we can see that there's a quadratic factor that can be simplified. The polynomial in the parentheses can be easily factored because it is of a special class of quadratics: half of the middle number squared is equal to the last number**.

$\displaystyle =4ax(x^2+2*(\underline{-6})+(\underline{-6})^2)$

$\displaystyle =4ax(x+(\underline{-6}))^2$

$\displaystyle =4ax(x-6)^2$

Which is our answer.

Remember, to check any factoring problem, one can expand the terms using the distributive property to see if the end result is the original polynonmial.

** In case there's some confusion about what I meant about the quadratic factor, consider this:

$\displaystyle x^2-\overbrace{12}^{middle}x+\overbrace{36}^{last}$  is our quadratic. half of the middle number $\displaystyle 1/2*\overbrace{-12}^{middle}$ equals $\displaystyle -6$. And $\displaystyle (-6)^2 = 36$ which is equal to the last term.

This whole process is similar to "completing the square". 

$\displaystyle 3c^2-39c+120$

Factor out the largest common quantity:

$\displaystyle 3(c^2-13c+40)$

Which two numbers can add/subtract to the middle term, but multiply to equal the last term?

$\displaystyle 3(c-8)(c-5)$

The product of negative 8 and negative 5 is positive 40. Their difference is also negative 13.

$\displaystyle 8x^2+32x+30$

Factor out the largest quantity common to all terms:

$\displaystyle 2(4x^2+16x+15)$

Factor the simplified quadratic:

$\displaystyle 2(2x+3)(2x+5)$

You need to use the sum of two cubes equation

$\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)$

To factor a polynomial that has a coefficient in front of the $\displaystyle x^2$ term, follow the steps below;

1) Once the equation is in standard form ($\displaystyle ax^2 +bx+c$) , multiply the $\displaystyle a$ term by the $\displaystyle c$ term

2) Find two factors of this term that give you the $\displaystyle b$ term

3) Re-write the polynomial with the original $\displaystyle b$ term expanded into the two factors

4) Factor by grouping

5) Distribute to check that the factorization is correct

You need to factor by grouping but the important step is to remember the difference of squares.

$\displaystyle a^2-b^2 = (a+b)(a-b)$