When we factor, we have to remember to check the signs in the trinomial. In this case, we have minus 4 and minus 12. That automatically tells us the signs in the factors must be opposite, one plus and one minus.

Next, we ask ourselves what are the factors of 12? We get 2 & 6, 1 & 12, and 4 & 3.

Then, we ask ourselves, which of these, when subtracted in a given order, will give us -4? The answer is 6 and 2! So we place these in the parentheses with the $\displaystyle x$'s that we know go there so it looks like

$\displaystyle \left ( x\ \ \ 2 \right )\left ( x\ \ \ 6 \right )$

Finally, we ask ourselves, what signs do we need to put in to get negative 4 and negative 12? We need a positive $\displaystyle 2x$ but a negative $\displaystyle 6x$ so we put the addition sign in with the 2 and the negative sign in with the 6! 

$\displaystyle \left ( x+2 \right )(x-6)$

When we factor, we have to remember to check the signs in the trinomial. In this case, we have plus 9 and plus 18. That automatically tells us the signs in the factors must be the same, two "+".

Next, we ask ourselves what are the factors of 18? We get 2 & 9, 3 & 6, and 1 & 18.

Then, we ask ourselves, which of these when added in a given order, will give us 9? The answer is 6 and 3! So we place these in the parentheses with the $\displaystyle x$'s that we know go there so it looks like

$\displaystyle (x\ \ \ \ 3)(x\ \ \ 6)$ 

Finally, we ask ourselves, what signs do we need to put in to get plus 9 and plus 18? We already answered that! Two "+" signs! So, the answer is: 

$\displaystyle (x+3)(x+6)$ 

When we factor, we have to remember to check the signs in the trinomial. In this case, we have minus 7 and plus 12. That automatically tells us the signs in the factors must be the same, both "-" signs.

Next, we ask ourselves what are the factors of 12? We get 2 & 6, 1 & 12, and 4 & 3.

Then, we ask ourselves, which of these when added/subtracted in a given order, will give us 7? The answer is 4 and 3! So we place these in the parentheses with the $\displaystyle x$'s that we know go there so it looks like

$\displaystyle (x - 4) (x - 3)$

When we factor, we have to remember to check the signs in the trinomial first. In this case, we have plus 11 and plus 14. That automatically tells us the signs in the factors must be the same, two "+" signs.

Next, we ask ourselves what are the factors of 14? We get 2 & 7 and 1 & 14.

Then, we need to account for the coefficient of $\displaystyle x^2$. That 2 needs to be in front of one of the $\displaystyle x$'s in the binomials, so our two binomials will look like this so far: 

$\displaystyle (2x + ?) (x + ? )$

Now we use those factors of 14 to fill in the blanks! We know it will be 2 and 7 because 14 plus 1 is more than 11! 2 plus 7 is less than 11. So, how will we get to 11? We know one of our constants will be multiplied by the 2 attached to the $\displaystyle x$ in the first binomial. Because multiplying 7 by 2 would get a number greater than 11, we know we put the 2 in the second binomial! That way the $\displaystyle 2x$ will multiply with the 2 and give us $\displaystyle 4x$, which, added to $\displaystyle 7x$ (the result of multiplying the other variable and constant) will give us $\displaystyle 11x$! So we know our answer will be: 

$\displaystyle (x + 2) (2x + 7)$

For the trinomial $\displaystyle \small \small \small x^{2}+Bx+36$ to be factorable, we would have to be able to find two integers with product 36 and sum $\displaystyle B$; that is, $\displaystyle B$ would have to be the sum of two integers whose product is 36.

Below are the five factor pairs of 36, with their sum listed next to them. $\displaystyle B$ must be one of those five sums to make the trinomial factorable.

1, 36: 37

2, 18: 20

3, 12: 15

4, 9: 13

6, 6: 12

Of the five choices, only 16 is not listed, so if $\displaystyle B = 16$, then the polynomial is prime.

To factor trinomials like this one, we need to do a reverse FOIL. In other words, we need to find two binomials that multiply together to yield $\displaystyle 7x^{2} - 11x - 6$.

Finding the "first" terms is relatively easy; they need to multiply together to give us $\displaystyle 7x^{2}$, and since $\displaystyle 7$ only has two factors, we know the terms must be $\displaystyle 7x$ and $\displaystyle x$. We now have $\displaystyle (7x$    $\displaystyle )(x$    $\displaystyle )$, and this is where it gets tricky.

The second terms must multiply together to give us $\displaystyle -6$, and they must also multiply with the first terms to give us a total result of $\displaystyle -11x$. Many terms fit the first criterion. $\displaystyle (-2)(3)$, $\displaystyle (-3)(2)$, $\displaystyle (-6)(1)$ and $\displaystyle (-1)(6)$ all multiply to yield $\displaystyle -6$. But the only way to also get the "$\displaystyle x$" terms to sum to $\displaystyle -11x$ is to use $\displaystyle (7x + 3)(x - 2)$. It's just like a puzzle!

Use the $\displaystyle ac$-method to split the middle term into the sum of two terms whose coefficients have sum $\displaystyle -31$ and product $\displaystyle 6*35 = 210$. These two numbers can be found, using trial and error, to be $\displaystyle -21$ and $\displaystyle -10$.

$\displaystyle -21*-10=210$ and $\displaystyle -21+(-10)=-31$

Now we know that $\displaystyle 6x^{2}-31x+35$ is equal to $\displaystyle 6x^{2}-10x-21x+35$.

Factor by grouping.

$\displaystyle (6x^{2}-10x)-(21x-35)$

$\displaystyle 2x(3x-5)-7(3x-5)$

$\displaystyle (2x-7)(3x-5)$

First, we note that the coefficients have an LCD of 3, so we can factor that out:

$\displaystyle 3(x^{2} - 15x - 32)$

We try to factor further by factoring quadratic trinomial $\displaystyle x^{2} - 15x - 32$. We are looking to factor it into two factors$\displaystyle (x+ ? )(x+?)$, where the question marks are to be replaced by two integers whose product is $\displaystyle -32$ and whose sum is $\displaystyle -15$. 

We need to look at the factor pairs of $\displaystyle -32$ in which the negative number has the greater absolute value, and see which one has sum $\displaystyle -15$:

$\displaystyle \begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-32& -31\\ 2,-16&-14 \\ 4,-8& -4 \end{matrix}$

None of these pairs have the desired sum, so $\displaystyle x^{2} - 15x - 32$ is prime. $\displaystyle 3(x^{2} - 15x - 32)$ is the complete factorization.

We are looking to factor this quadratic trinomial into two factors$\displaystyle (x+ ? )(x+?)$, where the question marks are to be replaced by two integers whose product is $\displaystyle -32$ and whose sum is $\displaystyle -15$. 

We need to look at the factor pairs of $\displaystyle -32$ in which the negative number has the greater absolute value and the sum is $\displaystyle -15$:

$\displaystyle \begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-32& -31\\ 2,-16&-14 \\ 4,-8& -4 \end{matrix}$

None of these pairs have the desired sum, so the polynomial is prime.

We are looking to factor this quadratic trinomial into two factors$\displaystyle (x+ ? )(x+?)$, where the question marks are to be replaced by two integers whose product is $\displaystyle -20$ and whose sum is $\displaystyle -9$. 

We need to look at the factor pairs of $\displaystyle -20$ in which the negative number has the greater absolute value, and see which one has sum $\displaystyle -9$:

$\displaystyle \begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-20& -19\\ 2,-10&-8 \\ 4,-5& -1 \end{matrix}$

None of these pairs have the desired sum, so the polynomial is prime.