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.