Factoring $2k^{2} + 7k + 6$ I know that $2k^{2} + 7k + 6 = (k+2)(2k+3)$ But I am not quite sure how to get there.
I get stuck trying to factor out the leading coefficient: $2(k^{2} + \frac{7}{2}k + \frac{6}{2})$, and then if I was lucky I would find two numbers
that multiply to $\frac{6}{2}$ and add to $\frac{7}{2}$. I know that it is one of the ways of easily factoring, but it obviously does not work in all cases.
To explain my method clearly, this is just an example of my method working:
If I had to factor $k^{2} + 7k + 10$ I would say $2*5 = 10$ and $2+5 = 7$ So my two terms are 2 and 5 such that the factoring becomes $(k+2)(k+5)$.
So my question is, how to factor when I end up with fractions after removing the leading coefficient?
 A: An easy way to solve the factoring problem is to recast it as a system of equations.  Because 2 is a prime number, you know that it's going to look like this:
$$2k^2+7k+6 = (2k+a)(k+b), $$
where $a,b$ are unknown.
Expand to get:
$$2k^2+7k+6 = 2k^2+(a+2b)k+ab.$$  Thus,
$$a+2b = 7,\quad ab = 6.$$
Now, play with factors of 6 (1,2,3,6).  Using a bit of trial and error should get you the answer.
A: I'm a fan of splitting the middle term. Given a quadratic $ax^2 + bx + c$, here are the steps for this procedure:

*

*Find two numbers, $r$ and $s$, that multiply to $ac$ and add to $b$

*Split $bx$ into the sum $rx + sx$, so $ax^2 + bx + c = ax^2 + rx + sx + c$

*Find the GCF of $ax^2 + rx$ and the GCF of $sx + c$, and factor those GCFs out separately

*There will be a common factor (it's whatever's in parentheses, next to the two GCFs). Factor that out, and you'll be done.

Here's what this looks like with your example. We have $2k^2 + 7k + 6$.
First, we find two numbers that multiply to $ac = 12$ and that add to $b = 7$. That's $3$ and $4$.
Then, rewrite $7k = 3k + 4k$ in the quadratic:
$$ 2k^2 + 3k + 4k + 6$$
The GCF of $2k^2 + 3k$ is $k$, so we'll factor that out of the first pair of terms, and for the second pair, the GCF is $2$, so that's what we factor out of the second pair of terms:
$$ k(2k+3) + 2(2k+3) $$
The common factor in both remaining terms is $2k+3$, so we factor that out of the two terms to end up at the answer:
$$ (2k+3)(k+2) $$
Note, it wouldn't make a difference if you had done $7k = 4k + 3k$ instead of $3k + 4k$, it just changes the GCFs a bit, but you'll end up in the same place:
$$ 2k^2 + 4k + 3k + 6 $$
This time, the GCF of $2k^2 + 4k$ is $2k$, so that's what we factor out of the first pair, and it's $3$ for the second pair:
$$ 2k(k + 2) + 3(k + 2) $$
Then, the common factor is $k+2$, so that's what we factor out to get to our final answer:
$$ (k+2)(2k+3) $$
A: This is how I teach factoring in this case. The method is called "Slide and Divide." Here are the steps.

*

*Given a quadratic $ax^2 + bx + c$, "slide" the coefficient $a$ to the end by means of multiplication, giving the new polynomial $x^2 + bx + ac.$


*Factor this new polynomial in the usual manner - find two numbers $u, v$ whose product is $ac$ and sum is $b$.


*Write $x^2 + bx + ac = (x+u)(x+v)$.


*Reintroduce $a$ by division. Divide both $u$ and $v$ and write $\left(x + \frac{u}{a}\right) \left(x + \frac{v}{a}\right)$. Reduce these as needed.


*If any of the fractions $\frac{u}{a}$ or $\frac{v}{a}$ are not integer, "slide" the denominator to the front of $x$.
Let's see this in action. First, write $2k^2 +7k +6$ as $k^2 +7k + 12.$ Then factor the usual way, i.e.
$$k^2 + 7k + 12 = (k+3)(k+4).$$
Now divide both constant terms by $2$ to yield
$$(k+3)(k+4) \to \left(k+ \frac{3}{2}\right) \left(k+\frac{4}{2}\right)$$
(note the two expressions are not equal). Now, $\frac{4}{2} = 2$, but $\frac{3}{2}$ does not reduce. So, we slide the $2$ from the first factor to the front of the $x$, giving
$$\left(k+ \frac{3}{2}\right) \left(k+ 2\right) \to (2k+3)(k+2)$$
which is exactly what we needed!
