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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

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.

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

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.

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- Feb 7, 2012

- 2,702

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\: + \: ??).$$\displaystyle x^4+2x^3-8x^2+18x+9$

this is what i tried,

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

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

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.

- Mar 22, 2013

- 573

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

\(\displaystyle \beta \alpha \lambda \alpha \rho \kappa \alpha\)

.

PS : When did I start signing like chisigma?

- Feb 9, 2012

- 33

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...

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

\(\displaystyle \beta \alpha \lambda \alpha \rho \kappa \alpha\)

.

PS : When did I start signing like chisigma?

\(\displaystyle \mu \pi \alpha \lambda \alpha \rho \kappa \alpha \)

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

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

If so, there is a "nice" factorization.

[tex]x^4 + 2x^3 - 8x^2 - 18x - 9[/tex]

. . [tex]=\;x^4 + 2x^3 + x^2 - 9x^2 - 18x - 9[/tex]

. . [tex]=\; x^2(x^2+2x+1) - 9(x^2 + 2x + 1)[/tex]

. . [tex]=\;(x^2+2x+1)(x^2-9)[/tex]

. . [tex]=\;(x+1)^2(x-3)(x+3)[/tex]