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# x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
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x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol  [#permalink]

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17 Sep 2018, 01:24
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45% (medium)

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59% (01:31) correct 41% (01:48) wrong based on 38 sessions

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[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 $$x^4+12x^3+49x^2+78x+40?$$

$$A. x+1$$
$$B. x+2$$
$$C. x+3$$
$$D. x+4$$
$$E. x+5$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" NUS School Moderator Joined: 18 Jul 2018 Posts: 1028 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink] ### Show Tags 17 Sep 2018, 02:34 1 Trying the options 1 by 1. Option 1: If x+1 is a factor then x = -1 should satisfy the equation $$x^4$$+12$$x^3$$+49$$x^2$$+78x+40 Equation becomes zero. Hence x+1 is a factor of $$x^4$$+12$$x^3$$+49$$x^2$$+78x+40. Similarly trying other options. Only x = -3 doesn't satisfy. We can conclude that x+3 is not factor of $$x^4$$+12$$x^3$$+49$$x^2$$+78x+40. C is the answer. _________________ Press +1 Kudos If my post helps! GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 937 Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink] ### Show Tags 17 Sep 2018, 08:33 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. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7740 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol [#permalink] ### Show Tags 19 Sep 2018, 02:12 2 => Assume $$x^4+px^3+qx^2+rx+s = (x+a)(x+b)(x+c)(x+d).$$ Then, $$s = abcd.$$ The constant terms $$a, b, c$$ and $$d$$ of the linear factors are factors of the constant term s of the original polynomial. Since $$3$$ is not a factor of $$40, x + 3$$ cannot be a factor of the original polynomial $$x^4+12x^3+49x^2+78x+40.$$ Therefore, C is the answer. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol   [#permalink] 19 Sep 2018, 02:12
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