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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

\(x^4+12x^3+49x^2+78x+40\) can be factored as a product of four linear polynomials with integer coefficients. Which of following cannot be a factor of \(x^4+12x^3+49x^2+78x+40?\)

\(A. x+1\)
\(B. x+2\)
\(C. x+3\)
\(D. x+4\)
\(E. x+5\)

\(?\,\,\,:\,\,\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40\,\,\, \ne \,\,\,\left( {{\text{altern}}{\text{.}}\,\,{\text{choice}}} \right) \cdot p\left( x \right)\)

\(\left( A \right)\,\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = \left( {x + 1} \right) \cdot p\left( x \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, - 1\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{root}}\,\,{\text{of}}\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = 0\)
\({\left( { - 1} \right)^4} + 12{\left( { - 1} \right)^3} + 49{\left( { - 1} \right)^2} + 78\left( { - 1} \right) + 40\mathop = \limits^? 0\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( A \right)\,\,{\text{refuted}}\)

\(\left( B \right)\,\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = \left( {x + 2} \right) \cdot p\left( x \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, - 2\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{root}}\,\,{\text{of}}\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = 0\)
\({\left( { - 2} \right)^4} + 12{\left( { - 2} \right)^3} + 49{\left( { - 2} \right)^2} + 78\left( { - 2} \right) + 40\mathop = \limits^? 0\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( B \right)\,\,{\text{refuted}}\)

\(\left( C \right)\,\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = \left( {x + 3} \right) \cdot p\left( x \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, - 3\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{root}}\,\,{\text{of}}\,\,{x^4} + 12{x^3} + 49{x^2} + 78x + 40 = 0\)
\({\left( { - 3} \right)^4} + 12{\left( { - 3} \right)^3} + 49{\left( { - 3} \right)^2} + 78\left( { - 3} \right) + 40\mathop = \limits^? 0\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( C \right)\,\,{\text{is}}\,\,{\text{the}}\,\,{\text{right}}\,\,{\text{answer}}\)


This solution follows the notations and rationale taught in the GMATH method.

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Fabio.
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
Can this concept get tested on GMAT ?
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
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pramodmundhra15 wrote:
Can this concept get tested on GMAT ?


Hi, pramodmundhra15 !

Why not? All the definitions and properties envolved in the question stem and in (at least some of the potential) solutions are in the GMAT quant section scope.

Besides that, proper solutions take 2min or less to the implemented, therefore I cannot see any reason to doubt this question could be in the real exam, for sure!

Good luck and success in your studies,
Fabio.
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
fskilnik i believe GMAT doesn't test beyond quadratic equations.
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
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pramodmundhra15 wrote:
fskilnik i believe GMAT doesn't test beyond quadratic equations.


Hi again, pramodmundhra15 .

Polynomials are part of the GMAT scope even for degrees greater than 2, mainly in "factoring issues" like this one.

I will not continue this discussion any further, though.

Best Regards,
Fabio.
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink]
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