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When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)?

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When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)?  [#permalink]

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New post 26 Sep 2018, 04:03
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A
B
C
D
E

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Question Stats:

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[Math Revolution GMAT math practice question]

When \(f(x)=ax^2+bx+c (a≠0)\), is \(x+1\) a factor of \(f(x)?\)

\(1) f(1)=0\)
\(2) f(-1)=0\)

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Re: When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)?  [#permalink]

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New post 26 Sep 2018, 04:20
If x+1 a factor of f(x) or otherwise when x equals -1, f(x) should be zero.
Clearly B is sufficient.

B is the answer

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Re: When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)?  [#permalink]

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New post 26 Sep 2018, 08:17
MathRevolution wrote:
[Math Revolution GMAT math practice question]

When \(f(x)=ax^2+bx+c (a≠0)\), is \(x+1\) a factor of \(f(x)?\)

\(1) f(1)=0\)
\(2) f(-1)=0\)

\(a \ne 0\)

\(a{x^2} + bx + c\,\,\mathop = \limits^? \,\,\left( {x + 1} \right) \cdot p\left( x \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,a{\left( { - 1} \right)^2} + b\left( { - 1} \right) + c\,\,\mathop = \limits^? \,\,0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\boxed{b\,\,\mathop = \limits^? \,\,a + c}\)

\(\left( 1 \right)\,\,\,a + b + c = 0\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {2,0, - 2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\left( * \right)\, \hfill \\
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {1,2, - 3} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\left( {**} \right)\,\, \hfill \\
\end{gathered} \right.\)

\(\left( * \right)\,\,\,2{x^2} - 2 = 2\left( {x + 1} \right)\left( {x - 1} \right) = \left( {x + 1} \right) \cdot p\left( x \right)\,\,\,,\,\,\,\,p\left( x \right) = 2\left( {x - 1} \right)\)

\(\left( {**} \right)\,\,{x^2} + 2x - 3 = \left( {x - 1} \right)\left( {x + 3} \right)\)


\(\left( 2 \right)\,\,a{\left( { - 1} \right)^2} + b\left( { - 1} \right) + c = 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


The correct answer is therefore (B).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)?  [#permalink]

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New post 27 Sep 2018, 23:38
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

\(x+1\) is a factor of \(f(x)\) precisely when \(f(x) = (x+1)(ax+b)\) for some numbers \(a\) and \(b\). This occurs when \(f(-1) = 0\), but not necessarily when \(f(1) = 0\).

Thus, only condition 2) is sufficient.

Therefore, B is the answer.
Answer: B

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: When f(x)=ax^2+bx+c (a≠0), is x+1 a factor of f(x)? &nbs [#permalink] 27 Sep 2018, 23:38
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