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

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New post 17 Sep 2018, 01:24
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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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Re: x^4+12x^3+49x^2+78x+40 can be factored as a product of four linear pol  [#permalink]

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New post 17 Sep 2018, 02:34
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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.
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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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New post 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.
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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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New post 19 Sep 2018, 02:12
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=>

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
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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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