Originally Posted by
Jhevon factor by grouping. there is a common 18x in the first two terms, and a common -9 in the last two
write as $\displaystyle 2^2 - (6x)^2$
now apply the formula for the difference of two squares factorization
begin by factoring out the common 6
now, what do you aee?
these are one of those two-way quadratic thingys.
here's a tip: forget the y for a while and treat it as $\displaystyle 12x^2 - 5x - 3$ and factor that. when finished, attach a y to the constant terms, and that will take care of the y's in your expression
same advice as the last problem
treat this as $\displaystyle (x^2)^2 - (x^2) - 30$
this is a quadratic in $\displaystyle x^2$, factor it as you would a quadratic. but instead of $\displaystyle x$ you are using $\displaystyle x^2$
if it is hard for you to visualize, let $\displaystyle y = x^2$, then you have $\displaystyle y^2 - y - 30$. factorize this, and then replace all the $\displaystyle y$'s with $\displaystyle x^2$'s