The other day, while helping my friend on homework, I was stuck on a problem. And by stuck I mean I didn't want to guess rational solutions and then synthetic to find if they worked because that's too main stream and I had already done that to change it from a 4th degree to a third.
So I just happened to come across a little trick which I'll detail in this note on how to factor Polynomials such as .
Let me begin by saying: This trick isn't the best for two reasons: It doesn't work all that often, and it may be a waste of time. +it's already kinda similar to RR and remainder theorem.
But on the other hand, when it works, it can be very helpful.
We begin by synthetic division on the polynomial Divide the entire polynomial by and we get
Move the 10 over to the other side
From here, we prime factorize 10 to .
Assuming x is rational, it's quite clear that x=1 or 2 or 5 since we have an x in the front of our equation. Thus we can plug in x= 1,2,5 to see which work. After doing so, we find that 2 and 5 both work.
Now, you might be asking "well how is this quicker". And the thing is that all you have to do is check that each binomial is a factor of 10.
This can become especially helpful when the constant term becomes quite large and you are stuck pluging in large numbers which when raised to the 5th power such as can become quite strenuous to do by hand.
Here's another example
Move the factors to the right.
A little intuition\manipulation yields that
Looking at our expression now gives is that
Now it's quite obvious that are roots just by looking at the equation.
Thus after dividing our equation by we get
This our answer is