Question

3. Determine the zeros of: f(x) = x4 – x3 + 7x2 – 9x – 18.

Answer 1

x=-1,2,-3i,3i

Answer 2

i already did it, but im left with 2 and -1+- sqrt-35/-2??? what is wrong?

Answer 3

It can be factored as (x-2)(x+1)(x^2+9)

Answer 4

Easy way to know that 1 or -1 is a s zero is by seeing if all the coefficients somehow add up to 0.
In this case, 1+1 +7+9 - 18 = 0. So -1 or 1 is a root.

Answer 5

ok but look im supposed to do like this:

Step 1: Use the Rational Root Theorem
Step 2: Use Descartes' Rule of Signs
Step 3: Use the Fundamental Theorem of Algebra
Step 4: Test the factors
Step 5: Write the depressed equation
Step 6: Find the zeros
Step 7: Look at the graph the function
Step 8: Confirm the zeros by substituting

Answer 6

and im left in step 6
with 2 and -1+-sqrt-35/-2 by using the quadratic formula!
because of the discriminant!

Answer 7

here's the table!

Answer 8

jeezus christ you shouldn't be memorizing steps

Answer 9

Step 1, wont even work if the constant term isn't rational.

Answer 10

Also how does the fundamental theorem of algebra help you solve anything? Its an existence theorem not a constructive theorem.

Answer 11

In addition the proofs of these techniques are pretty advanced, I wouldn't be inclined to use techniques you can't understand, but thats just me.

Answer 12

i know right! but they want us to do things this way!
I dont know what to do!!??

Answer 13

i could factorize, i know but that;s out of " the steps"

Answer 14

and they take points out :(

Answer 15

So what do you need help with factoring, or finding the roots

Answer 16

Do you actually have to learn the mathematics, or do you just want to get it done?

Answer 17

Jack17
(x-2)(x+1)(x^2+9) is great and I know x^2+9 will result in x+-3i
no, Iw ant to learn the mathematics, no one spends his whiole summer learning!!

Answer 18

i want to understand thus thing
if not i would just have copied from yahoo or here!

Answer 19

sooo????

Answer 20

I would try to find rational roots if any, then use polynomial division progressively with Factor theorem. This also helps to narrow down the possible roots.

Answer 21

ok thanks!

Answer 22

now, is there any way it can be solved by completing the square or using the quadratic formula?

Answer 23

because that's the only thing i can turn in in my assignment, base on the discriminant i got!

Answer 24

which is -35 is that right?

Answer 25

Another helpful tip is roots come in complex conjugate pairs if the coefficients of the polynomial are real.
Completing the square or quad. formula only works once you narrow by poly division into quadratic factors.
Sometimes you get irreducible quadratic factors and that means there are non-real complex roots.

Answer 26

so if i get \[-1\pm \sqrt{-35}/ -2\]

Answer 27

that means it has non real complex roots?

Answer 28

is this for the same question? if you have a negative number in the discriminant then it has non real complex roots. You would rewrite it as sqrt(35) * i

Answer 29

thta's what i was looking for, so it is\[-1+\sqrt{35*i}/-2\]

Answer 30

um the i never goes in the sqrt! keep the i outside of it. You could write it as i sqrt(35) if you like. But that's not the answer for the question you posted. A different question?

Answer 31

Determine the zeros of: f(x) = x4 – x3 + 7x2 – 9x – 18 <----- right?

Answer 32

can u do it by completing the square please?

Answer 33

sorry the quadratic formula, because i did it and that's what i got!

Answer 34

Yep I might take you through this step by step:
First test any rational roots. 1 and -1 are easy
P(1) = (1)^4 - (1)^3 +7(1)^2 -9(1) - 18 which is not 0.
P(-1) =... = 0 so x + 1 must be a factor by factor Theorem.

Answer 35

Dividing the original poly by x+1 yields x^3 -2x^2 -9x - 18. Now we need to look for more roots. 1 did not work before, so try other factors of -18, say 2.
P(2) also yields 0 so x + 2 must also be a root.

Answer 36

Sorry x - 2 is a root not x + 2

Answer 37

x - 2 being a factor, dividing x^3 -2x^2 -9x - 18 by this now gives x^2 + 9

Answer 38

So solutions to x^2 + 9 = 0 gives you your complex roots.
(x^2 + 9) = (x - 3i)(x + 3i)
Our roots are then -1,2, 3i and -3i.
There was no need to use quad. formula or complete the square.
If you did use quadratic formula with x^2 + 9 = 0, you'd get b^2 - 4ac being (0)^2 - 4(1)(9) = -36 or (6i)^2. Then dividing by a = 2, gives you plus or minus of 3i

Answer 39

\[-(b)\pm \sqrt{b ^{2}-4ac}/2(a)\] <------------ quadratic formula

Answer 40

whoops yes
i meant a = 1, 2a = 2

Answer 41

whattt?