+6251 find all the common factors and bring them to the front of the expression.
In this case 10y is a common factor.
$$30y^3-50y=10y\left(3y^2-5\right)$$
it should be fine to stop there, but if you want to go nuts you can do this.
$$30y^3-50y=30y\left(y^2-\frac 5 3\right)=30y\left(y-\sqrt{\frac 5 3}\right)\left(y+\sqrt{\frac 5 3}\right)$$
+6251 find all the common factors and bring them to the front of the expression.
In this case 10y is a common factor.
$$30y^3-50y=10y\left(3y^2-5\right)$$
it should be fine to stop there, but if you want to go nuts you can do this.
$$30y^3-50y=30y\left(y^2-\frac 5 3\right)=30y\left(y-\sqrt{\frac 5 3}\right)\left(y+\sqrt{\frac 5 3}\right)$$
+129847 I used to use that second form of factoring, too, Rom. It drove my teachers crazy!! I asked them why a radical wasn't just as valid as an integer in a factored expression since they were both "real" numbers......They use to say, "Because you can't do that." I told them, "But, I just did!!"
