How many unique roots of the equation (x^3 + 3x^2 - 9x + 5)/(x - 1) =

Question

GMATBusters’ Quant Quiz Question -8

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How many unique roots of the equation (x^3 + 3x^2 - 9x + 5)/(x - 1) = 0?

A. 0
B. 1
C. 2
D. 3
E. None of the above

GMATBusters · Answer

Official Solution: 
 (x^3 + 3x^2 - 9x + 5)/(x-1)=0
 first point to note that x-1 is in denominator, so x cannot be 1 
(x^2+4x-5)(x-1)/(x-1)=0
x^2+4x-5=0
(x-1)(x+5)= 0
As x is not equal to 1, 
x= -5
only one solution
Hence answer B

Raxit85 · Answer

How many unique roots of the equation (x^3+3x^2-9x+5)/(x-1) = 0?

(x^3+3x^2-9x+5)/(x-1) = 0
(x^3+ 3 x^2-9x+ 5) , where product is 5 and sum/difference is 3
Factor of 5 are 1 & 5 but we need one more factor as it's cubic equation. So, 1,5,& -3 that can make the sum of 3. 
So, (x-1) (x-5) (x+3)/(x-1) = 0
(x-5) (x+3) = 0
x = 5/-3

Ans. C

A. 0
B. 1
C. 2
D. 3
E. None of the above

ashwini2k6jha · Answer

x^3-x^2+4x^2-4x-5x-+5/x-1=0
(x^2+4x-5)(x-1)/x-1=0
x^2+4x-5=0
x=1,-5
as x=not1
Hence answer B

Archit3110 · Answer

at x= 1 ,-5
we get (x^3+3x^2-9x+5)/(x-1) = 0
IMO B ; 1, -5 as x=1 not possible

How many unique roots of the equation (x^3+3x^2-9x+5)/(x-1) = 0?
A. 0
B. 1
C. 2
D. 3
E. None of the above

uc26 · Answer

For every negative number, we get a negative result for LHS. Example x= -2.
For x=0, we get a result of -5 on the LHS.
For every positive number, we get a positive result for LHS. Example x=3.
Thus the LHS can never equal 0 which means the given equation has no unique roots. 
Answer: A.

Kinshook · Answer

Asked: How many unique roots of the equation  (x^3+3x^2-9x+5)/(x-1) = 0?

x^3 + 3x^2 -9x + 5 = 0  &  x 
eq 1 
If x=1;  x^3 + 3x^2 -9x + 5 = 1+ 3 -9 + 5 = 0 ; x=1 is a root & (x-1) is a factor of  x^3 + 3x^2 -9x + 5 
 x^3 + 3x^2 -9x + 5 = x^2(x-1) +4x(x-1) -5(x-1) = (x-1)(x^2+4x-5) = (x-1)(x+5)(x-1) = (x-1)^2(x+5) 
Since  x 
eq 1 ; x = -5 is the only root of the equation  (x^3+3x^2-9x+5)/(x-1) = 0

IMO B

K1oxHvs_zHE

freedom128 · Answer

Note that x=1 is NOT a valid solution for equation (x^3 + 3x^2 - 9x + 5)/(x - 1) = 0. The divisor (x-1) must not be zero

(x^3 + 3x^2 - 9x + 5)/(x - 1) = 0
--> (x-1)(x^2 +4x -5)/(x-1) = 0
--> (x-1)(x-1)(x+5)/(x-1) = 0
--> x=1 (Not applicable) or x=-5 (Yes)

Therefore, there is only 1 (one) unique root of such equation.

FINAL ANSWER IS (B) 1 only

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shameekv1989 · Answer

How many unique roots of the equation (x^3 + 3x^2 - 9x + 5)/(x - 1) = 0?

A. 0
B. 1
C. 2
D. 3
E. None of the above

\frac{(x^3 + 3x^2 - 9x + 5)}{(x-1)}  =  \frac{(x^3 - x^2 + 4x^2 - 9x + 5)}{(x-1)}  =  \frac{(x^2(x-1) + (4x^2-9x+5))}{(x-1)}  =  \frac{(x^2(x-1) + (x-1)(4x-5))}{(x-1)}  =  \frac{(x^2+4x-5)(x-1)}{(x-1)}

x^2+4x-5 = (x+5)(x-1)

But x can not be equal to 1

Hence x = -5 is the only root of the equation

Answer - B

Abhishek009 · Answer

\frac{(x^3 + 3x^2 - 9x + 5)}{(x - 1)} = 0

Or,  \frac{(x−1)(x−1)(x+5)}{(x - 1)} = 0

So,  (x−1)(x+5) = 0

Hence unique value of  x = -5  , Answer must be (B)

omema · Answer

Why X=1 is not possible

Bunuel · Answer

When x = 1, then  \frac{(x^3 + 3x^2 - 9x + 5)}{(x - 1) }= \frac{(x^3 + 3x^2 - 9x + 5)}{0} , which is undefined. Recall that division by 0 is not allowed.

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How many unique roots of the equation (x^3 + 3x^2 - 9x + 5)/(x - 1) =

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