Question

Find all the zeros of the eq. 2x^4-5x^3+53x^2-125x+75=0

Answer 1

This one can be factorized, so you will need to use the following tools:

Descartes rule of signs,
factorization theorem,
fundamental theorem of algebra

Are any or some of these famililar to you?
Also, this may indicate that you or your teacher have a typo in the other problem. Please re-check for typos on the other problem.

Answer 2

To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.

Answer 3

no I dnt know those methods

Answer 4

Descartes rule of signs tells us that we have either 4 real positive roots, or two real positive roots and two complex roots, or 4 complex roots.

Answer 5

ok

Answer 6

http://en.wikipedia.org/wiki/Descartes'_rule_of_signs
http://en.wikipedia.org/wiki/Factor_theorem
http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
Since you have been missing school for at least one day, it may well be advisable to catch up with some of these topics. Hope you got notes from your friends though.

Answer 7

So is (x-1) a factor to the equation?

Answer 8

yes

Answer 9

So you can do a synthetic division, or a long division to reduce the quartic to a cubic, i.e. find (2*x^4-5*x^3+53*x^2+(-125)*x+75)/(x-1)=?

Answer 10

2x^3-3x^2+50x-75

Answer 11

Good, now we can rule out the third case (4 complex roots).

Answer 12

ok

Answer 13

Using the factor theorem, you will need to try factors of the form
(ax+b) where a=1 or 2 (from the leading coefficient 2), and b=+/- all possible factors of 75, which are 1,3,5,15,25,75, i.e. 2*2*6=24 possibilities.
Once you find that, you can again do the synthetic division or long division to reduce the cubic to a quadratic.

Answer 14

what do I divide it by

Answer 15

When you find another root (i.e. another factor) of the cubic.

Answer 16

oh ok, can I use the calculator for that

Answer 17

Sorry to bust in here, but \(2x^3-3x^2+50x-75\) is relatively easily factored by grouping, and thus we can skip the whole guess and check portion of finding roots.

Answer 18

In fact, Descartes rule of sign says we don't have to try factors of the form (a+b), only those of the form (a-b) because all real roots are positive.

Answer 19

So that reduces to 12 possibilities only.

Answer 20

Very true, thanks KingGeorge!

Answer 21

You group like so \[2x^3-3x^2+50x-75=x^2(2x-3)+25(2x-3)\]Can evy15 finish the factoring from here?

Answer 22

@evy15 are you able to complete the solution from here?

Answer 23

can you guide me to what to do next

Answer 24

ok hold on

Answer 25

x=i+-5 and 3/2

Answer 26

i mean =-5i

Answer 27

Good job, (+\- 5i).
Now you have all 4 zeroes!

Answer 28

Can you enumerate the four zeroes?

Answer 29

so its 1,+/-5i, and 3/2?

Answer 30

or-1

Answer 31

Excellent! There you are! Not -1. Your list is good.

Answer 32

Big thanks to KingGeorge who cut the work in half!

Answer 33

yes thank you guys so much

Answer 34

yw! :)

Answer 35

As an addendum, it's possible to factor the original function by grouping if you're clever.
\[\begin{aligned}
2x^4-5x^3+53x^2-125x+75&=2x^4-5x^3+3x^2+50x^2-125x+75 \\
&=x^2(2x^2-5x+3)+25(2x^2-5x+3) \\
&=(x^2+25)(2x^2-5x+3)\\
&=(x^2+25)(x-1)(2x-3)
\end{aligned}\]

Answer 36

But this is not an obvious, nor the easiest, way to do it.

Answer 37

Yep, I will look more carefully for these cases in the future.

Answer 38

can i do the same with x^5-3x^4-24x^3-72x^2+12x=12

Answer 39

sorry wrote the wrong one

Answer 40

x^5-3x^4-24x^3-72x^2-25x+75=0

Answer 41

That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.

Answer 42

Your first group will be \[x^4(x-3).\]Knowing this, can you give me the rest of the factorization?

Answer 43

x^4(x-3)-24x^2(x-3)-25(x-3)

Answer 44

Perfect. And can you finish the rest of the factorization?

Answer 45

would it be (x^4-24x^2-25) and (x-3)?

Answer 46

Right again. Now you just need to factor \(x^4-24x^2-25\). You can do this by pretending it's a quadratic equation, and not a quartic.

So how does \(x^2-24x-25\) factor?

Answer 47

x-25 and x+1

Answer 48

Perfect. So that means \(x^4-24x^2-25=(x^2-25)(x^2+1)\). Hence, \[x^5-3x^4-24x^3-72x^2-25x+75=(x^2-25)(x^2+1)(x-3)\]From here, you just have some sums/differences of squares, and should be able to figure out the roots.

I've got to go now, but you seem to be catching the hang of this!