Normally you would move everything to the left. In this case that's already done. So if I understand it correctly I have a trinomial equation with two variables at this point.
So I need to have to have two terms since there are two different unknown factors. So then I find factors of 32: 16, 8, 4, 2.
My question is why out of those factors do I use -4 and -8? Is it simply because when combined they equal -12x?
Normally you would move everything to the left. In this case that's already done. So if I understand it correctly I have a trinomial equation with two variables at this point. - no you have one variable - 'x'
So I need to have to have two terms since there are two different unknown factors. So then I find factors of 32: 16, 8, 4, 2.
My question is why out of those factors do I use -4 and -8? Is it simply because when combined the equal -12x? ............. Yes
It helps a lot to think about multiplying polynomials when you are doing factoring!
Yes, looking at x2−12x+32 (which has the single variable, x. Surely you were not thinking of "x" and x2 as different variables?) you should think about the possible integer factors of 32. And, having done that, recognise that (-8)+ (-4)= -12 so the factors are (x- 8)(x- 4).
The reason that works is because (x−8)(x−4)=x(x−4)−8(x−4) ("distributive law", distributing x- 4 over the two parts of x- 8) which is equal to x2−4x−8x−32 (distributing the x and -4 over the x and 8) and then the two "middle terms" add: -4x- 8x= -12x. That is why we look for factors of 32 that add to -8.
(By the way, most of the time we cannot find integer factors of the "constant term" that sum to the coefficient of x. Most polynomials cannot be factored with integer coefficients. For example, if the polynomial were x2−12x+31, the only factors of 31 are 1, -1 31, and -31. And none of those add to -12. Of course, we could solve the equation x2−12x+31=0 by "completing the square" or using the "quadratic formula". We would find that the two roots are 6+5 and 6−5. That tells us that we can factor this polynomial as (x−(6+5))(x−(6−5))=((x−6)−5)((x−6)+5). As a check, that last is the "product of sum and difference"- it is equal to (x−6)2−(5)2=x2−12x+36−5=x2−12x+31.
Theoretically we could have tried to factor the original polynomial with square roots but there are simply too many possible factors that way.)
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