factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$
I tried with $5x(5x+y)-6y^2$. I'm stuck here. 
I can't continue.  
 A: Forget about the $y$ for a second. 
If you want to factor $25x^2 + 5x - 6$, look for two numbers whose product is $25 \cdot -6 = -150$ and whose sum is $5$. It takes just a moment to see that the two numbers are $15$ and $-10$. Then:
\begin{align*}
25 x^2 + 5x - 6 &= 25 x^2 + (15x - 10x) - 6 \\ &= (25 x^2 + 15x) - (10x + 6) \\ &= 5x(5x + 3) - 2(5x + 3) \\& = (5x-2)(5x+3).\end{align*}
With the $y$ included the process is nearly identical:
\begin{align*}
25 x^2 + 5xy - 6 y^2&= 25 x^2 + (15xy - 10xy) - 6y^2 \\ &= (25 x^2 + 15xy) - (10xy + 6y^2) \\ &= 5x(5x + 3y) - 2y(5x + 3y) \\& = (5x-2y)(5x+3y).\end{align*}
A: Hint: Factor $25x^2+5x-6$ first.
A: The trick here is to manipulate the expression $25x^2+5xy-6y^2$. Try the following:
\begin{align}
25x^2+5xy-6y^2 &= 25x^2+15xy-10xy-6y^2\tag{manipulate}\\[0.5em]
               &= 5x(5x+3y)-2y(5x+3y)\tag{factoring}\\[0.5em]
               &= (5x-2y)(5x+3y)\tag{group}
\end{align}
Is this clear now?
A: Instead of finding all of the factors of -150, lets do this a bit more intelligently. Notice that 25 and 5 are both multiple of 5. This gives us the opportunity to substitute $t=5x$
$$25x^2+5xy-6y^2=(5x)^2+(5x)y-6y^2=t^2+ty-6y^2=\dots$$
After factoring, just substitute the $5x$ back in.
A: Or just use $abc$ formula, with $a = 25$, $b = 5y$ and $c = -6y^2$, and find that terms disappear like snow before the sun:
$$ \frac{-5y ^+_- \sqrt{625y^2} }{2 \times 25}$$
So $5x = -3y$ or $5x = 2y$, etc.
A: let me expand on my comment of using quadratic formula to factor an expression like this. of course, you need to have the quadratic formula that
$x = \dfrac{-b + \pm \sqrt{b^2 - 4ac}}{2a}$ are the solutions of 
$ax^2 + bx + c = 0.$ 
we will look at first the equation $25x^2 +5xy - 6y^2 = 0$ later on we will see how we can use the solutions to factor the expression. matching $25x^2 +5xy - 6y^2 = 0$ with $ax^2 + bx + c = 0$ we can identify $a = 25, b = 5y, c = -6y^2.$ first the discriminant $b^2 - 4ac = (5y)^2 - 4*25*(-6y^2) = 625y^2 = (25y)^2$
using the formula $x = \dfrac{-5y \pm 25y}{50} = -\dfrac{3y}{5}, \dfrac{2y}{5}$ 
we can turn this into two equations into the factors 
$k(x+\frac{3y}{5})(x - \frac{2y}{5}) = 25x^2 +5xy - 6y^2 = 0 $ matching the coeffcient of $x^2$ fixes $k = 25.$
what do we have now is $$25(x+\frac{3y}{5})(x - \frac{2y}{5})= (5x + 3y)(5x - 2y)= 25x^2 +5xy - 6y^2 = 0 $$ 
from the last equalities drop leftmost and the rightmost expressions and you have factored $25x^2 +5xy - 6y^2.$
A: The answer is $(5x-2y)(5x+3y)$.
A: Let me add some explanations on Umberto P.'s answer, which is the same as that given in
http://www.adamlchan.com/math/topics/ac_factoring/index.php
cited by The Chaz 2.o.
Assume that
$$Ax^2+Bxy+Cy^2=(Px+Qy)(Rx+Sy)$$
Then
$$Ax^2+Bxy+Cy^2=(PR)x^2+(PS+QR)xy+(QS)y^2$$
Let's first find $(PS)$ and $(QR)$.
Step 1: Calculate $AC$.
Then observe that
$$(PS)(QR)=AC$$
$$PS+QR=B$$
Step 2: Find two numbers whose product is $AC$ and whose sum is $B$.
After obtaining $PS$ and $QR$,
Step 3: Rewrite the original equation as $(PR)x^2+(PS)xy+(QR)xy+(QS)y^2$
Step 4: Grouping the terms and factorize them 
$$[(PR)x^2+(PS)xy]+[(QR)xy+(QS)y^2]=Px(Rx+Sy)+Qy(Rx+Sy)$$
Step 5: Applying distributive law
$$Px(Rx+Sy)+Qy(Rx+Sy)=(Px+Qy)(Rx+Sy)$$
Applying to this problem:
Step 1: $25\times(-6)=-150$
Step 2: Find two numbers whose product is -150 and whose sum is 5 --> 15 and -10
Step 3: Write it as $25x^2+15xy-10xy-6y^2$
Step 4: $(25x^2+15xy)+(-10xy-6y^2)=5x(5x+3)-2y(5x+3y)$
Step 5: $5x(5x+3)-2y(5x+3y)=(5x-2y)(5x+3y)$
