How do you factor completely #x^2-12x+35#?
We need to find two numbers that add up to
as our factored form, and to find the zeros, take the opposite sign to get:
So what my math teacher used to teach me is that the place where the
However, the numbers have to be the same, so what do we do?
We first list all the factors of
#1 * 35#
#5 * 7#
#7 * 5#
#35 * 1#
And then the negatives (wink-wink it's going to be negatives...)
#-1 * -35#
#-5 * -7#
#-7 * -5#
#-35 * -1#
And then the one negative one positive... and so on. However, did you noticed that the middle number is negative?
A positive times positive can only have a positive result. That is eliminated. If it is a negative time a positive, then it can only be:
#-35 + 1 = -34#
#-1 + 35 = 34#
#-7 + 5 = -2#
#-5 + 7 = 2#
That are the only possible answers. But none of them are
So the only way is negative times a negative. And the only combination that results in a
Voila, you have your answer. And of course, you put it with the
#x * x#
#-7 * -5#
Hope this helps!! :)
Find the 2 factors of 35 that also add to -12
Because our equation is already in the form
Begin by finding the factors that multiply to 35 and add to -12.
Personally, if I don't see the factors after a few seconds I will begin listing ALL the factors of the "c" term and begin fiddling with the signs of the individual factors until I find something that works.
We then take those factors and rewrite our equation in the following form
These are the factors that you will be looking for. To check our work, simply distribute to undo the factoring process.