Factoring the polynomial $5c^2 - 52c + 20$ I was trying to find the factors of that expression above:
$5c^2-52c+20.$
The solution is
$(5c−2)(c−10).$
I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?
 A: $$5c^2-52c+20=0$$
$$5c^2-50c-2c+20=0\\5c(c-10)-2(c-10)=0$$
$$(5c-2)(c-10)=0$$
or you can use quadratic formula $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Here $a=5,b=-52,c=20$
A: Break the expression into the groups
$$\left(5c^2-50c\right)+\left(-2c+20\right)$$
Factor out $5c$ from $5c^2-50$, which yields $5c\left(c-10\right)$. Now factor out $-2$ from $-2c+20$ and we get $-2\left(c-10\right)$
Factor out the common term $\left(c-10\right)$
$$\left(c-10\right)\left(5c-2\right)$$
A: There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):


*

*First write $\left(\dfrac{ax\phantom{+4}}{\phantom{5}}\right)
\left(\dfrac{ax\phantom{+4}}{\phantom{5}}\right).$

*Find the product $ac$, including sign.

*Find the prime factorization of $ac$ using the factor tree. 

*Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $\sqrt{ac}$.

*Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.

*Write $\left(\dfrac{ax+s}{\phantom{5}}\right)\left(\dfrac{ax+t}{\phantom{5}}\right).$

*Divide each of these binomials by its own GCF:
$\left(\dfrac{ax+s}{\operatorname{gcf}(a,s)}\right)
\left(\dfrac{ax+t}{\operatorname{gcf}(a,t)}\right).$ Check that
$$\operatorname{gcf}(a,s)\cdot\operatorname{gcf}(a,t)=a.$$ 

A: $$\begin{align}
5c^2 - 52c + 20 &= \frac{5}{5}\cdot\big(5c^2 - 52c + 20\big)\\
&= \frac{25c^2 - 260c + 100}{5}\\
&= \frac{(5c)^2 -52(5c) + 100}{5}\\
&= \frac{t^2 -52t + 100}{5}\text{ where } t=5c\\
\end{align}$$
Can you take it from there?
