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

- anonymous

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

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- mathmate

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.

- mathmate

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

- anonymous

no I dnt know those methods

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- mathmate

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.

- anonymous

ok

- mathmate

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.

- mathmate

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

- anonymous

yes

- mathmate

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)=?

- anonymous

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

- mathmate

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

- anonymous

ok

- mathmate

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.

- anonymous

what do I divide it by

- mathmate

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

- anonymous

oh ok, can I use the calculator for that

- KingGeorge

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.

- mathmate

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.

- mathmate

So that reduces to 12 possibilities only.

- mathmate

Very true, thanks KingGeorge!

- KingGeorge

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

- mathmate

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

- anonymous

can you guide me to what to do next

- anonymous

ok hold on

- anonymous

x=i+-5 and 3/2

- anonymous

i mean =-5i

- mathmate

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

- mathmate

Can you enumerate the four zeroes?

- anonymous

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

- anonymous

or-1

- mathmate

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

- mathmate

Big thanks to KingGeorge who cut the work in half!

- anonymous

yes thank you guys so much

- mathmate

yw! :)

- KingGeorge

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

- KingGeorge

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

- mathmate

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

- anonymous

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

- anonymous

sorry wrote the wrong one

- anonymous

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

- KingGeorge

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

- KingGeorge

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

- anonymous

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

- KingGeorge

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

- anonymous

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

- KingGeorge

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?

- anonymous

x-25 and x+1

- KingGeorge

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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.