My parents had me move, and I am in a new school and they're currently learning factoring. I need to learn this in order to catch up with my new class. I have a very hard time with Algebra, so please explain in an easy step by step process for me. I think if I get help with this one problem I can figure out the others.
The equation I need to factor is:
2x^2+x-6
Before you start "factoring" polynomials you should have learned to
multiply them!
\(\displaystyle (ax+ b)(cx+ d)= acx^2+ (ad+ bc)x+ bd\)
So you are looking for numbers, a, b, c, and d, such that ac= 2, ad+ bc= 1, and bc= -6.
Now, that would be nearly
impossible except that when we talk about "factoring" in this sense, we
mean "factoring with
integer coefficients".
You proceed by "educated" trial and error.
"2" can only be factored, with
integers, as (1)(2) so
either a=1 and c= 2
or a= 2 and c= 1.
Similarly 6 can be factored as (1)(6), (6)(1), (2)(3), or (3)(2). Also the factors of -6 must have opposite signs.
So we have quite a few, but still only "a few" possible values for b and d:
b= 1, d= -6 or b= -6, d= 1, or b= 2, d= -3, or b= -3, d=2, or b= -1, d= 6, or b= 6, d= -1, or b= -2, d= 3, or b= 3, d= -2.
That is 2 possible cases for a and c and 8 cases for b and d so (2)(8)= 16 cases altogether. We determine which is correct (if any-
most polynomials
don't factor with integer coefficients) by checking ad+ bc= 1.
The "educated" part is recognizing that we don't have to try all of them. Obviously, if have b or d equal to 6 or -6, we wont be able to get "ad+ bc= 1" so we should try just the "3""2" factors for -6. One thing that should jump out at you is that ad+ bc= (2)(2)+ (1)(-3)= 1 so a= 2, d= 2, b= -3, c= 1 should work.
Check by multiplying (ax+ b)(cx+ d)= (2x- 3)(x+ 2) to see what you get
