jramirez23 said:
I am having a very difficult time learning how to factor ax[sup:33tqulyc]2[/sup:33tqulyc] + bx + c.
I am especially having problems with the following equation:
4x[sup:33tqulyc]2[/sup:33tqulyc] + 27x + 35.
I am extremely clueless; I have even tried some websites and I still need help.
Here's a method I always use for factoring trinomials of the form ax[sup:33tqulyc]2[/sup:33tqulyc] + bx + c where "a" is something other than 1.
Multiply "a" * "c". In your example, multiply 4 * 35, to get 140.
Now, look for two numbers whose product is 140, and whose SUM is "b," or 27 in your example.
7 and 20 will work, since 7*20 = 140, and 7 + 20 = 27.
Rewrite the middle term as 7x + 20x:
4x[sup:33tqulyc]2[/sup:33tqulyc] + 7x + 20x + 35
Factor by grouping. Remove a common factor of x from the first two terms, and a common factor of 5 from the last two terms:
x(4x + 7) + 5(4x + 7)
Remove the common factor of (4x + 7):
(4x + 7)(x + 5)
There's the factorization you're looking for. This may sound complicated, but it really isn't. It takes MUCH longer to explain it than it does to
do it. I like this method because it does not involve any trial-and-error, and
always works on any
factorable trinomial of the form ax[sup:33tqulyc]2[/sup:33tqulyc] + bx + c.