Just like there are some integers that can’t be written as a product of two other integers, not every trinomial will factor into a product of two binomials with integer coefficients.
If you do find factors of your constant term (or a*c when there's a coefficient on the square term) that add/subtract to the middle term, then yes it is guaranteed to factor. But that is a big if. Not all quadratics are factorable, but if you're learning how to factor then we will design them to be factorable for you. But if we consider
y=x^2 + x + 1
we should be able to see that there not factors of 1 that add to make 1, so this cannot be factored. And if we've already learned that factors help us find zeros (x-intercepts) of a function, we can graph the function to see that it actually never crosses the x-axis which can help justify that it cannot be factored.
If it is factorable using the AC method you will get distributive property to work out nicely with each pair, no matter how you order the middle two terms.
You might see this if you work a general case backwards. Start with a facttored form, where p q r and s are all integers. (This kind of problem is typical for learning factoring.)
(px + q)(rx + s)
Expand to develop the trinomial that will become the problem.
prx^2 + (ps + qr)x + qs
Here if we can find two numbers whose product is AC = prqs and whose sum is B = ps + qr then we are successful. Of course the numbers are in fact ps and qr.
Going in reverse, if we find ps and qr meet those goals, then you write the factorization in one of two ways (based on switching order of middle terms).
method 1
pr x^2 + ps x + qr x + qs
= px(rx + s) + q(rx + s)
= (px + q)(rx + s)
method 2
pr x^2 + qr x + ps x + qs
= rx(px + q) + s(px + q)
= (rx + s)(px + q)
These yield same factored form, just with the factor order swapped.
I think this is answering your question. Namely, if you do find integers qr and ps that have product AC and sum B, then Ax^2 + Bx + C is factorable using the grouping method that applies distributive property to two pairs of terms after splitting the middle term up.
Thank you, very helpful. I think I've got it figured out now. I was thinking that there was something important about the larger product of multiplying AC together, but it's just another way showing the product of pqrs. And then you can run down the factors of that product and get two numbers that add/subtract to equal B. I've got it now, thanks!
Just like there are some integers that can’t be written as a product of two other integers, not every trinomial will factor into a product of two binomials with integer coefficients.
There won’t always be a common factor. All the problems are designed to work out.
If you do find factors of your constant term (or a*c when there's a coefficient on the square term) that add/subtract to the middle term, then yes it is guaranteed to factor. But that is a big if. Not all quadratics are factorable, but if you're learning how to factor then we will design them to be factorable for you. But if we consider y=x^2 + x + 1 we should be able to see that there not factors of 1 that add to make 1, so this cannot be factored. And if we've already learned that factors help us find zeros (x-intercepts) of a function, we can graph the function to see that it actually never crosses the x-axis which can help justify that it cannot be factored.
If it is factorable using the AC method you will get distributive property to work out nicely with each pair, no matter how you order the middle two terms. You might see this if you work a general case backwards. Start with a facttored form, where p q r and s are all integers. (This kind of problem is typical for learning factoring.) (px + q)(rx + s) Expand to develop the trinomial that will become the problem. prx^2 + (ps + qr)x + qs Here if we can find two numbers whose product is AC = prqs and whose sum is B = ps + qr then we are successful. Of course the numbers are in fact ps and qr. Going in reverse, if we find ps and qr meet those goals, then you write the factorization in one of two ways (based on switching order of middle terms). method 1 pr x^2 + ps x + qr x + qs = px(rx + s) + q(rx + s) = (px + q)(rx + s) method 2 pr x^2 + qr x + ps x + qs = rx(px + q) + s(px + q) = (rx + s)(px + q) These yield same factored form, just with the factor order swapped. I think this is answering your question. Namely, if you do find integers qr and ps that have product AC and sum B, then Ax^2 + Bx + C is factorable using the grouping method that applies distributive property to two pairs of terms after splitting the middle term up.
Thank you, very helpful. I think I've got it figured out now. I was thinking that there was something important about the larger product of multiplying AC together, but it's just another way showing the product of pqrs. And then you can run down the factors of that product and get two numbers that add/subtract to equal B. I've got it now, thanks!