Factoring polynomials with a 2nd degree coefficient greater than $1$ I'm learning how to factor polynomials, but I'm having a hard time understanding the approach when the 2nd degree coefficient is greater than $1$.
For example, when I begin to factor $12k^4 + 22k^3 - 70k^2$, I first break it down to $2k^2(6k^2 + 11k - 35)$.
I would think that I'd want to find two numbers that sum up to $11$ and have a product of $-35$, but instead I'm told we need to multiply $-35$ by $6$ so that I now have to find two numbers that sum up to $11$ and have a product of $-210$.
Can anyone help me understand why $-35$ is multiplied by the coefficient $6$?  Why isn't $11k$ also multiplied by $6$?

To split the linear term of $6k^2 + 11k - 35$, we must find two numbers with product $6 \cdot -35 = -210$ and sum $11$.  They are $21$ and $-10$.  Hence,
\begin{align*}
6x^2 + 11k - 35 & = \color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35} && \text{split the linear term}\\
                & = 3x(2x + 7) - 5(2x + 7) && \text{factor by grouping}\\
                & = (3x - 5)(2x + 7) && \text{extract the common factor}
\end{align*} 
Notice that in the expression $\color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35}$, the product of the quadratic and constant coefficient is equal to the product of the two linear coefficients whose sum is the linear coefficient of $6x^2 + 11k - 35$, that is, 
$$(\color{blue}{6})(\color{blue}{-35}) = (\color{green}{21})(\color{green}{-10}) = -210$$
Let's examine the general case.  Suppose $ax^2 + bx + c$ factors with respect to the rationals as $(rx + s)(tx + u)$.  Then 
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
              & = rx(tx + u) + s(tx + u)\\
              & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{green}{su}\\
& = \color{blue}{rt}x^2 + (\color{green}{ru} + \color{green}{st})x + \color{blue}{su}
\end{align*}
Matching coefficients, we have $a = \color{blue}{rt}$, $b = \color{green}{ru + st}$, and $c = \color{blue}{su}$.  Notice that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is $b$, that is, 
$$ac = (\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$
Thus, to split the linear term, we must find two numbers whose product is $ac$ and whose sum is $b$.
