Qwerty90 Group Title Question about how to solve: Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18. I am solving this problem using all of the polynomial ways such as decartes rule, fundamental therom and rational root. But it seems when I solve it that way, I either come up one answer short or a completely different answer from what I see everyone else get. How would you solve this and what is the answer? one year ago one year ago

1. ZeHanz

RRT says: Rational solutions can only be factors of 18 divided by factors of the coefficient of x^4. Rational roots therefore have the form:$\pm 18, \pm9, \pm6, \pm3, \pm2, \pm1$If you try (begin with the simplest, of course, so 1 or -1), you get -1 as a root. You can now write f(x) as (x+1)(x³ ..........). To get the 3rd degree part, you can do a long division or a synthetic division. As soon as you have found it, you can do this trick again: I think 2 is a root as well (just try it), so you can factor out x-2 to get (x+1)(x-2)(x²........). Now you can esily check if there are more rational or even real roots.

2. ZeHanz

Could it be that you are not familiar with synthetic division? It's a real handy trick!

3. Qwerty90

No I know how to do synthetic division, its just after that where everything get messy.

4. ZeHanz

OK, so you've found: f(x)=(x+1)(x-2)(x²+9). As you can see, the last factor x²+9 has no real roots, only complex ones: 3i and -3i.

5. Qwerty90

So, basically after you found which zeros is part of the function you list it in parentheses as the conjugate. Now my last question is where did you get the x+9 from? i know that 9 is the constant but what about the x^2?

6. ZeHanz

After the first synthetic division, I got (x+1)(x³-2x²+9x-18). Then the number 2 is also a zero, so I divided x³-2x²+9x-18 by x-2. That gave me x²+9