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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
Posts: 6826
GMAT 1: 760 Q51 V42
GPA: 3.82
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, 00:24
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Difficulty:

35% (medium)

Question Stats:

59% (01:11) correct 41% (01:25) wrong based on 29 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 $$x^4+12x^3+49x^2+78x+40?$$

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

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 18 Jul 2018 Posts: 591 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, 01:34 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. _________________ If you are not badly hurt, you don't learn. If you don't learn, you don't grow. If you don't grow, you don't live. If you don't live, you don't know your worth. If you don't know your worth, then what's the point? GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 629 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, 07:33 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: 6826 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, 01:12 1 => 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$149 for 3 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 &nbs [#permalink] 19 Sep 2018, 01:12
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