MHB Factoring 4th degree polynomial.

[paulmdrdo1]

$\displaystyle x^4+2x^3-8x^2+18x+9$

this is what i tried,

$x^4+2x^3-8x^2 = (x^2+4x)(x^2-2x)$

then,

$a(x^2+4x)+b(x^2-2x)=18x$

where ab=9

did i set up my solution correctly? can you tell me where I'm wrong.

 

[MarkFL]

I don't find a nice factorization for this quartic polynomial. Are you sure it is given correctly?

 

[Opalg]

$\displaystyle x^4+2x^3-8x^2+18x+9$

this is what i tried,

$x^4+2x^3-8x^2 = (x^2+4x)(x^2-2x)$

I am pretty sure that MarkFL is correct and that you have copied the question wrongly. My guess is that the $+\:9$ at the end should be $-\:9$. If so, then your partial factorisation $x^4+2x^3-8x^2 = (x^2+4x)(x^2-2x)$ is a very good step in the right direction. All you have to do is to add a constant to each factor: $x^4+2x^3-8x^2 + 18x +9 = (x^2+4x\: +\: ??)(x^2-2x\: + \: ??).$

 

[Prove It]

$\displaystyle x^4+2x^3-8x^2+18x+9$

this is what i tried,

$x^4+2x^3-8x^2 = (x^2+4x)(x^2-2x)$

then,

$a(x^2+4x)+b(x^2-2x)=18x$

where ab=9

did i set up my solution correctly? can you tell me where I'm wrong.

If it was \displaystyle \begin{align*} x^4 + 2x^3 + 8x^2 + 18x - 9 \end{align*} it would factorise nicely by grouping. Are you sure that's not what it should be?

 

[mathbalarka]

Indeed, the polynomial cannot be factored in Z[x]. Unless you clarify, there is going to be a lot of trouble identifying the typo. Here's a hint of what I am talking about :

You give the quartic polynomial (1, 2, -8, 18, 9) which is provably irreducible over Z[x]. The factorizable polynomials can be found by "tweaking" the coefficients a bit, i.e., (1, 2, -8, 18, -9), (1, 2, 8, 18, -9), (1, 2, -8, -18, -9), (1, -2, -8, -18, -9), (1, -2, -8, 18, -9), (1, -2, 8, -18, -9), etc are all provably splittable over Z[x], and I've barely covered the signing problems

$$\beta \alpha \lambda \alpha \rho \kappa \alpha$$
.

PS : When did I start signing like chisigma?

 

[agentredlum]

Indeed, the polynomial cannot be factored in Z[x]. Unless you clarify, there is going to be a lot of trouble identifying the typo. Here's a hint of what I am talking about :

You give the quartic polynomial (1, 2, -8, 18, 9) which is provably irreducible over Z[x]. The factorizable polynomials can be found by "tweaking" the coefficients a bit, i.e., (1, 2, -8, 18, -9), (1, 2, 8, 18, -9), (1, 2, -8, -18, -9), (1, -2, -8, -18, -9), (1, -2, -8, 18, -9), (1, -2, 8, -18, -9), etc are all provably splittable over Z[x], and I've barely covered the signing problems

$$\beta \alpha \lambda \alpha \rho \kappa \alpha$$
.

PS : When did I start signing like chisigma?

Greetings and salutations my friend, Just as a note , and i realize Greek is not your native language , pronounced in Greek we would say valarka , there is no single letter in Greek that has a B (bee) sound , the beta is actually called 'vee - ta' and is pronounced exactly like English 'v', to get the 'B' sound a Greek would have to use two letters and write mu - pi , like this...

$$ \mu \pi \alpha \lambda \alpha \rho \kappa \alpha $$

The only reason i know this is because I'm part Greek.

:)

 

[soroban]

Hello, paulmdrdo!

Could the polynomial be: .x^4 + 2x^3 - 8x^2 \,{\color{red}-}\,18x \,{\color{red}-}\,9\,?

If so, there is a "nice" factorization.

x^4 + 2x^3 - 8x^2 - 18x - 9

. . =\;x^4 + 2x^3 + x^2 - 9x^2 - 18x - 9

. . =\; x^2(x^2+2x+1) - 9(x^2 + 2x + 1)

. . =\;(x^2+2x+1)(x^2-9)

. . =\;(x+1)^2(x-3)(x+3)