Question

how do you find the zeros of f(x)=x^4+5x^3+10x^2+20x+24?

Answer 1

wolfram alpha

Answer 2

or try some negative numbers that divide \(24\)
the zeros clearly cannot be positive right?

Answer 3

For this we have to use something called the rational zeros theorem. What that requires is that we take the factors of the constant number, 24 in this case, and divide it by all the factors of the leading coefficient (the number in front of the highest power of x), which is 1 in this case. When we do this, our results are all of our POSSIBLE zeros of the polynomial. So we set it up like this:

\[\frac{ \pm 24 }{ \pm 1}\]
doing this means all of our possible zeros are
\[\pm1, \pm2, \pm3, \pm4, \pm6, \pm12, \pm24 \]Now this is a lot of possiblities, but when we pick one it can help us get an idea if we need to guess higher or lower. This make sense so far?

Answer 4

Descartes rules? am I supposed to find factors of 24 and 1?

Answer 5

Correct :3 So basically what I have done above.

Answer 6

so far so good

Answer 7

Okay, awesome. Now I have a general idea what the answer will be, so I am going to choose -3 as our number to check. Once we choose a number, we do synthetic division and see if the answer turns out to have no remainder. You know how to do synthetic division?

Answer 8

yes i do, i got 390 so this cant be a root

Answer 9

Ah. You must have done something wrong, I do show it to be a root. Do you want to see my work?

Answer 10

yes please

Answer 11

oh i got it i put in a +3 by mistake