Factorizing -λ³+7λ+6=0

[Rob K]

Hi guys,

Need some help on something I didn't know I didn't know.

How do you factorize this?
-λ³+7λ+6=0

I assumed this?

-λ(λ²-7) + 6 = 0
λ = 6
λ = ±√7

But that's wrong as the answers in the book are -2, -1, and 3.

Can someone give me a little guidance on this, I have only ever done factorization where one of them is already know.

Kind Regards

Rob K

 

[e^(i Pi)+1=0]

First, I would just change the lambdas to X to make the problem look more familiar. To find one of the zeros, graph it if you're allowed, or use the rational root test to find one. Then, divide like normal, ect..

http://www.purplemath.com/modules/rtnlroot.htm

 

[Rob K]

Actually, I am working out an eigenvalue problem hence the Lamdas. I was kind of expecting to find a way to do this algebraically. If there were an x² or λ² term in there, I would be able to do it, but the lack of it has stumped me.

 

[jedishrfu]

I suppose you could solve it like you do for quadratics:

rewriting it to x^3 - 7x -6 = 0

and then looking at the factors of 6: 1, 2, 3, 6

then plug each one into the equation to find a root realizing you must try both +- versions.
From there you have one factor (x +- ??) that you can use to divide
the original polynomial by to get a quadratic that you can solve.

 

[micromass]

Actually, I am working out an eigenvalue problem hence the Lamdas. I was kind of expecting to find a way to do this algebraically. If there were an x² or λ² term in there, I would be able to do it, but the lack of it has stumped me.

So if I write it as x³ + 0x² - 7x -6 = 0, you'll be able to solve it?

 

[Borek]

x^3 - 7x -6 = 0

Makes me think of Cardano's method.

 

[e^(i Pi)+1=0]

Actually, I am working out an eigenvalue problem hence the Lamdas. I was kind of expecting to find a way to do this algebraically. If there were an x² or λ² term in there, I would be able to do it, but the lack of it has stumped me.

How is the rational root test and algebraic/synthetic long division not algebraic?

 

[Mark44]

How do you factorize this?
-λ³+7λ+6=0

I assumed this?

-λ(λ²-7) + 6 = 0
λ = 6
λ = ±√7

No offense meant, but that's not even remotely close. What you're saying here is something like, if a*b = 6, then a = 6 or b = 6.

If two numbers multiply to some other number, there's not much you can say about either of the two numbers, unless the third number happens to be 0.

In other words, writing the equation as
λ(λ²-7) = 6
is pretty much a nonstarter.

How is the rational root test and algebraic/synthetic long division not algebraic?

This is definitely the way to go.

 

[Rob K]

Apologies, I missed the comment about the rational root test. I don't know it, but now I know what it is called, I can learn it, thank you.

After some reading I didn't realize that this is actually a fairly complex problem to do manually. I assumed as quadratics are easily solved by either factorization or the equation method, that other polynomials are as easy.

Regards

Rob

 

[ehild]

I suppose you know that a polynomial p(x)=∑a_ix_i of order n can be factorised in terms of the roots x₁, x₂, ...x_n of the equation p(x)=0:
p(x)=a_n(x-x₁)(x-x₂)...(x-x_n).

Multiplying the factors, the product has the constant term a_nx₁ x₂...x_n which must be the same as a₀, the constant term of the original polynomial. When all coefficients are integer rational roots can exist. They can be obtained in the form (divisor of a₀)/(divisor of a_n)

You can write your polynomial as λ³-7λ -6, so a_n=1 and a₀=-6 The integer divisors of 6 are ±1, ±2, ±3, ±6. Try to find a root among these numbers. What do you get when subbing -1 for λ, for example? If you find one root, x₁, divide the polynomial by (x-x₁) and you get a quadratics easy to factorise, or try other possible root.

ehild

 

[ehild]

Makes me think of Cardano's method.

It is certainly effective but I'd rather death ...

ehild