**Question from Rachel:**

I’m on 4.2 of Algebra 2 at the moment. I made it through the lesson and understood it until I got to the last section, titled “Finding Rational and Imaginary Zeros.” I understood most of that with the exception of a couple of things. Here are my questions.

#1) Why, when using synthetic division to find the zeros of the polynomial does the book only use the positive version of the possible numbers? For example, +/-1 and +/-5 were the possible rational zeros in Example 5. In the division table, they only used a positive 1. I know that was because they found the zero right off, but from what they’ve been doing, they would only have used positive 1 and positive 5. Could you tell me why?

#2) In the matched problem after example 5, the polynomial looks like this:

P(x) = X^4 + 4X^3 + 10X^2 + 12X + 5.

When you find the possible rational zeros (+/- 1 and +/- 5), and plug them into the synthetic division table, none of them (as positive numbers – like the book has been using) turn out to be zeros!

On the 5 for example, 5 times 1 is 5. Added to 4, it becomes 9. 9 times 5 is 40. Added to 10 it becomes 50. 5 times 50 is 250. Added to 12, it becomes 262. 5 times 262 is 1310. Added to 5, it is 1315.

Now I know that that’s not supposed to be happening. I just don’t know why it is! It seems like there needs to be some negatives somewhere to cancel things out. I could use some help figuring it out.

Thanks for all your help! I appreciate it!

**Answer from Dr. Callahan:**

The book only used the positive numbers in Ex 5 because they just started there and they happened to work. Nothing else, just they started with the easiest.

In the other, 5 does not work because it is not a zero. Just because you have POSSIBLE zeros, does not mean they work. You are testing values to see which work and which do not. When they do, you have factored it down.

See http://www.wolframalpha.com/input/?i=factor+X%5E4+%2B+4X%5E3+%2B+10X%5E2+%2B+12X+%2B+5

This shows that you have ONE zero here at x=-1. Also, note that you cannot factor anymore since there are no other REAL zeros.

Remember, the whole thing here is just a tool to find placed where the equation =0.