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

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

Chethan92 · Answer

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

Posted from my mobile device

fskilnik · Answer

a 
e 0

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

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

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

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

\left( 2 ight)\,\,a{\left( { - 1} ight)^2} + b\left( { - 1} ight) + c = 0\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{	ext{YES}}} ightangle

The correct answer is therefore (B).

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

Regards, 
Fabio.

MathRevolution · Answer

=>

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.

QuantMadeEasy · Answer

According to factor theorem: a polynomial f(x) has a factor (x-a) only if f(a) = 0 
Hence, x+1 is factor of f(x) only if f(-1) = 0

B is correct

GMATGuruNY · Answer

(x + 1) is a factor of  ax^2+bx+c  if the quadratic is equal to 0 when x=-1.
To illustrate:
 x^2 + 4x + 3 = 0 
 (x+1)(x+3) = 0 
 x=-1  or  x=-3 
When  x=-1 ,  x^2 + 4x + 3 = 1 - 4 + 3 = 0 .

Question stem, rephrased:
Is the function equal to 0 when x=-1?

Statement 1:
f(1) = 0 --> When x=1, the value of the function is 0.
No information about the value of the function when x=-1.
INSUFFICIENT.

Statement 2:
f(-1) = 0 --> When x=-1, the value of the function is 0.
Thus, the answer to the question stem is YES.
SUFFICIENT.

B

MBAB123 · Answer

Hi  GMATGuruNY , Could you please explain how we get the highlighted part? I am not sure I follow the illustration.

GMATGuruNY · Answer

To solve a quadratic, we generally set it equal to 0 and determine its factors, as follows:
 x^2 - 2x = 1 5
 x^2 - 2x - 15 = 0 
 (x-5)(x+3) = 0 
x=5 or x=-3

The FACTORS of the quadratic are (x-5) and (x+3).
The ROOTS are x=5 and x=-3.

When a root is plugged into its corresponding factor, the factor has a value of 0.
If we plug x=5 into (x-5), we get:
5-5 = 0
Since a root causes one of the factors to have a value of 0, it also causes the quadratic itself to have a value of 0.
If we plug x=5 into  x^2-2x-15 , we get:
 5^2-2*5-15=0

If we know a factor of a quadratic, we can determine the value of one root -- a value that will cause both the factor and thus the quadratic itself to have a value of 0.
If (x-10) is a factor  ax^2+bx+c , then x=10 is a root, implying that x=10 will cause both (x-10) and  ax^2+bx+c  to have a value of 0.
If (x+7) is a factor  ax^2+bx+c , then x=-7 is a root, implying that x=-7 will cause both (x+7) and  ax^2+bx+c to have a value of 0.
If (2x-3) is a factor  ax^2+bx+c , then x=3/2 is a root, implying that x=3/2 will cause both (2x-3) and  ax^2+bx+c  to have a value of 0.

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

For (x+1) to be a factor of  ax^2+bx+c , x=-1 must be a root, implying that x=-1 will cause both (x+1) and  ax^2+bx+c  to have a value of 0.
In other words:
If x=-1 causes the function to have a value of 0, then (x+1) will be a factor of the quadratic, and the answer to the question stem will be YES.
Rephrasing the question stem, we get:
 Is the function equal to 0 when x=-1?

Clear?

MBAB123 · Answer

To solve a quadratic, we generally set it equal to 0 and determine its factors, as follows:
 x^2 - 2x = 1 5
 x^2 - 2x - 15 = 0 
 (x-5)(x+3) = 0 
x=5 or x=-3

The FACTORS of the quadratic are (x-5) and (x+3).
The ROOTS are x=5 and x=-3.

When a root is plugged into a factor, the factor has a value of 0.
If we plug x=5 into (x-5), we get:
5-5 = 0
Since a root causes one of the factors to have a value of 0, it also causes the quadratic itself to have a value of 0.
If we plug x=5 into  x^2-2x-15 , we get:
 5^2-2*5-15=0

If we know a factor of a quadratic, we can determine the value of one root -- a value that will cause both the factor and thus the quadratic itself to have a value of 0.
If (x-10) is a factor  ax^2+bx+c , then x=10 is a root, implying that x=10 will cause both (x-10) and  ax^2+bx+c  to have a value of 0.
If (x+7) is a factor  ax^2+bx+c , then x=-7 is a root, implying that x=-7 will cause both (x+7) and  ax^2+bx+c to have a value of 0.
If (2x-3) is a factor  ax^2+bx+c , then x=3/2 is a root, implying that x=3/2 will cause both (2x-3) and  ax^2+bx+c  to have a value of 0.

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

For (x+1) to be a factor of  ax^2+bx+c , x=-1 must be a root, implying that x=-1 will cause both (x+1) and  ax^2+bx+c  to have a value of 0.
In other words:
If x=-1 causes the function to have a value of 0, then (x+1) will be a factor of the quadratic, and the answer to the question stem will be YES.
Rephrasing the question stem, we get:
 Is the function equal to 0 when x=-1?

Clear?

Crystal Clear. Thank you sir!

bumpbot · Answer

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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)?

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