How to determine the zeros of this equation?

Question

Hi :) I'm an Algebra 2 student and I'm having a bit of trouble answering this question.
Determine the zeros of f(x)=x^4-x^3+7x^2-9x-18.
I have almost all of the work to this problem done. Here's my work:
Step 1: Use the Rational root Theorem:
factors of p: +-1,+-2,+-3,+-6,+-9,+-18
factors of q: +-1
factors of p/q (simplified) : +-1,+-2,+-3,+-6,+-9,+-18

Step 2:Use Descartes' Rule of Signs:
Ater using this I found these possibilities:
3 positive real zeros, 1 negative real zero and 1 complex real zero.
or
0 positive real zeros, 1 negative real zero and 3 complex real zero.

Step 4: Test the factors:
I used synthetic division to text which of the factors of p (from step 1) would create a remainder of 0.
The results were that only 2 and -1 created remainders of 0.

Step 5: Write the depressed equations.
After using synthetis division i cam up with these equations:
2: x^3+x^2+9x+9 and -1: x^3-2x^2+9x-18

Now my trouble is, how do you simplify those two equations from step 5 in order to solve for x? I don't know if I factor or use the quadratic formula. Also, sorry for not typing out everything. I have a very sore wrist and it really hurts to type. You don't have to give me the answers, but just please guide me in the right direction. Thank you very much!!

? · Accepted Answer

So close!

You found f(x) =  (x^3+x^2+9x+9)(x-2)

The next thing to do is not to divide f(x) by (x+1).

What you need to do next is divide x^3+x^2+9x+9 by (x+1).

You should get

f(x) = (x^2 + 9)(x+1)(x-2)

and go on from there.

Good guy · Answer

Hello,

Good, you have made much work !

Now you know that 2 and -1 are roots of your polynomial
It means that this one can be written :
(x-2)(x^3+x^2+9x+9)
and that the quadrinomial (x^3+x^2+9x+9) is still divisible by (x+1)
Use again synthetic division !
You find that (x^3+x^2+9x+9) = (x+1)(x^2+9)

And the initial polynomial is equal to :
(x-2)(x+1)(x^2+9)

x^2+9 has two complex roots which are 3i and -3i

Therefore 2 real zeros (2 and -1) and 2 complex zeros (3i and -3i)...

Hope it helped

Bye !

   



With these two quadrinomials, you start again to

Anonymous · Answer

as quickly as you have factored the equation you have (2x - a million)*(x + 5) = 0 From this equation you be responsive to that the two 2x - a million could be 0 or x + 5 could be 0. so which you clean up the two a sort of aspects for 0 to locate the zeroes. For the 2d you basically clean up for x 64x^2 - 25 = 0 x^2 = 25/sixty 4 x= +5/8 or x = - 5/8

Anonymous · Answer

As you had found x-2 & x+1 can divide the polynomial from step 4, you
may do the following:

1 - 1 + 7 - 9 -18 | 2
...+2 + 2+18+18|
----------------------|
1 +1 + 9 +9........|-1
....- 1+ 0 - 9........|
----------------------|
1 + 0 + 9
So, what you get finally is
x^2+9 a quadratic.
We may write
x^4-x^3+7x^2-9x-18=0=>
(x-2)(x+1)(x^2+9)=0=>
x=2,-1,+/-3i
are the zeros.

? · Answer

If you found a root there is no point in looking another root in the same equation. You should use the equation you came up for finding other roots. So I guess you'll need to "find" one root again. It's probably faster to "find" -1 in x^3+x^2+9x+9. After that your equation will be like this: 
(x-2)(x+1)(some kinda quadratic)

Anonymous · Answer

You have two roots (zeros) already.    x = -1   and   x = 2
So the quadratic  (x+1)(x-2) is a factor of the original quartic.
So x^2 - x - 2 is a factor of it.
Divide the quartic by the quadratic and get the resulting quadratic.
Then solve this quadratic.   You will then have all 4 zeros of the original quartic.

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How to determine the zeros of this equation?

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