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#### bergausstein

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- Jul 30, 2013

- 191

any hints on how to start this problem?

$12x^4+19x^3-26x^2-61x-28$

$12x^4+19x^3-26x^2-61x-28$

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- Thread starter
- #1

- Jul 30, 2013

- 191

any hints on how to start this problem?

$12x^4+19x^3-26x^2-61x-28$

$12x^4+19x^3-26x^2-61x-28$

Last edited:

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- Feb 7, 2012

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Is that $12x^2$ perhaps a typo for $12x^4$?any hints on how to start this problem?

$12x^2+19x^3-26x^2-61x-28$

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- Jul 30, 2013

- 191

yes that's 12x^4. sorry.

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- Feb 7, 2012

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- Feb 15, 2012

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The roots may not be INTEGERS, as the leading term's coefficient is not 1.....

- Aug 7, 2013

- 21

i will use dorobostikerlines method.,

$12(x^2-1)^2+19(x^2-1)(x+1)-21(x+1)^2$

i'll let you continue.

$12(x^2-1)^2+19(x^2-1)(x+1)-21(x+1)^2$

i'll let you continue.

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- #7

- Feb 7, 2012

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True, but I like an easy life, so I look for the simplest possible solutions first.The roots may not be INTEGERS, as the leading term's coefficient is not 1.....

Any hints on how to start this problem?

$\text{Factor: }\:f(x) \:=\:12x^4+19x^3-26x^2-61x-28$

[tex]\text{We find that }f(\text{-}1) \,=\,0.[/tex]

[tex]\text{Hence, }x+1\text{ is a factor.}[/tex]

[tex]\text{Long division: }\:f(x) \:=\: (x+1)\underbrace{(12x^3 + 7x^2 - 33x - 28)}_{g(x)}[/tex]

[tex]\text{We find that }g(\text{-}1) \,=\,0.[/tex]

[tex]\text{Hence, }x+1\text{ is a factor.}[/tex]

[tex]\text{Long division: }\:g(x) \:=\: (x+1)(12x^2-5x - 28)[/tex]

[tex]\text{Can you finish it?}[/tex]

- Aug 30, 2012

- 1,143

I've never heard of this method and google comes up with nothing. Can you give us a quick run-down?i will use dorobostikerlines method.,

$12(x^2-1)^2+19(x^2-1)(x+1)-21(x+1)^2$

i'll let you continue.

-Dan

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- Jul 30, 2013

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