Problem Solving

[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

Max@Math Revolution 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

The product of all the numbers not coefficients of X must equal 40. 3 is not a factor of 40 C, X+3

Max@Math Revolution 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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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