Math gurus...I need precalculus help

[EKS]

This problem came home as homework today.

 

Algebraically solve 6 - 3x - 4x² - 2x³ = 0

 

I know from graphing it on the computer that it has one real root which is irrational.  I know how to find irrational roots by factoring stuff out until I get a quadratic.  But I can't do that with this problem.  

 

So, how does one do this problem? 

 

(I honestly think it might be a typo.  Either that or the teacher is trying to torture the students.  With this particular teacher, either is possible :cursing: )

 

Thanks!

[Janeway]

Google this... and you get tons of answers including the steps.

Edited September 28, 2016 by Janeway

[Janeway]

In fact, just click on my link about and you will see the Google search.

[Arcadia]

My kid suggest using Vieta Formula with roots as a, b+ci, b-ci.

Three equations and three unknowns.

 

He is busy reading at the library or I would have ask him to work it out for you.

 

ETA:

Using Vieta's formula gave me back the original equation. I'll ask my kid to try later.

 

ETA:

Kid tried with Vieta and got back the same equation too.

Edited September 29, 2016 by Arcadia

[Caroline]

This problem came home as homework today.

 

Algebraically solve 6 - 3x - 4x² - 2x³ = 0

 

I know from graphing it on the computer that it has one real root which is irrational. I know how to find irrational roots by factoring stuff out until I get a quadratic. But I can't do that with this problem.

 

So, how does one do this problem?

 

(I honestly think it might be a typo. Either that or the teacher is trying to torture the students. With this particular teacher, either is possible :cursing: )

 

Thanks!

If you change one sign, it is factorable by grouping. I vote typo.

[SparklyUnicorn]

does seem like a sign typo

 

My instructor made 3 sign errors today rendering three problems not doable the way she wanted them done...LOL.  So it happens!!

 

 

[EKS]

If you change one sign, it is factorable by grouping. I vote typo.

 

This is what I'm leaning toward as well.

[EKS]

Google this... and you get tons of answers including the steps.

 

I have...

 

Everything I found has reducing (or whatever it's called) it to a quadratic and then using the quadratic formula.

Edited September 28, 2016 by EKS

[EKS]

My kid suggest using Vieta Formula with roots as a, b+ci, b-ci.

Three equations and three unknowns.

 

He is busy reading at the library or I would have ask him to work it out for you.

 

I'll look this up.  Thanks!

[Guest]

Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms:

 

3(2-x)-2x^2(2-x)=0

 

Then factor out the (2-x) from each of the two remaining terms:

 

(2-x)(3-2x^2)=0

 

Since it is equal to zero, then each of the two multiplied factors can be set to zero:

 

2-x=0 and 3-2x^2=0

 

You can probably solve it from here.

 

OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL

Edited September 29, 2016 by Kinsa

[Caroline]

Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms:

 

3(2-x)-2x^2(2-x)=0

 

Then factor out the (2-x) from each of the two remaining terms:

 

(2-x)(3-2x^2)=0

 

Since it is equal to zero, then each of the two multiplied factors can be set to zero:

 

2-x=0 and 3-2x^2=0

 

You can probably solve it from here.

 

OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL

That's why I thought it was probably a typo.

[EKS]

Start by factoring a 3 out of the first two terms, and a -2x^2 out of the next two terms:

 

3(2-x)-2x^2(2-x)=0

 

Then factor out the (2-x) from each of the two remaining terms:

 

(2-x)(3-2x^2)=0

 

Since it is equal to zero, then each of the two multiplied factors can be set to zero:

 

2-x=0 and 3-2x^2=0

 

You can probably solve it from here.

 

OOPS! WAIT! I think I have a sign error... don't listen to me. I shouldn't do math while cooking dinner! LOL

 

That's what I did at first too!

[MarkT]

Wolfram gives

 

x ~  0.80121   real root 

x ~ -1.4006 +- 1.3352i  complex roots

 

you can display it in exact form

 

which is a nasty looking solution - good luck - but it can be done

Edited September 29, 2016 by MarkT

[EKS]

Wolfram gives

 

x ~  0.80121   real root 

x ~ -1.4006 +- 1.3352i  complex roots

 

you can display it in exact form

 

which is a nasty looking solution - good luck - but it can be done

 

Yup.  Ugh.

 

In talking to the kid when he got home, I'm fairly sure a negative sign was left off, because they had actually done some work with factoring by grouping in class.

 

Thanks to everyone for your help!