If f(x) = x^3 - 4x^2 +x -4, then f(x) = 0 for how many real values of

Question

If  f(x) = x^{3} - 4x^{2} + x - 4 , then  f(x) = 0  for how many real values of x?

A. 0
B. 1
C. 2
D. 3
E. 4

Kinshook · Accepted Answer

JessicaJ

Asked: If  f(x) = x^{3} - 4x^{2} + x - 4 , then  f(x) = 0  for how many real values of x?

f(x) = x^3 - 4x^2 + x - 4 = 0
f(x) = x^2(x-4) + (x-4) = 0
f(x) = (x^2 + 1)(x-4)=0
x = 4 is the only real solution

IMO B

MartyMurray · Answer

Explanation

If f(x) =  x^{3}  -  4x^{2}  + x - 4, then f(x) = 0 for how many real values of x?

We need to determine how many real values of  x  there are such that  x^{3} - 4x^{2}+ x - 4 = 0 .

We can find the roots by factoring.

To factor, we can notice that  x^{3} - 4x^{2}  and  x - 4  both have  x - 4  as a factor. So, we can divide  x^{3} - 4x^{2}+ x - 4  by  x -4  and get the following:

(x^{2})(x - 4)+ (1)(x - 4) = 0

(x^{2}+ 1)(x - 4) = 0

For  (x^{2}+ 1)(x - 4)  to equal  0 , either  x^{2}+ 1  or  x - 4  must equal  0 .

There are no real values of  x  such that  x^{2}+ 1 = 0  since, for all real values of  x ,  x^{2}+ 1 > 0

There is one real value of  x  such that  x - 4 = 0 , which is 4.

So, there is one real value of x such that  (x^{2})(x - 4)+ (1)(x - 4) = 0

Correct Answer   B

BrushMyQuant · Answer

f(x) = x^{3} - 4x^{2} + x - 4 = 0  
=>  x*x^{2} - 4x^{2} + x - 4 = 0 
 =>  x^{2} * ( x - 4 ) + 1*( x - 4 )= 0 
=>  (x^{2} + 1 )* ( x - 4 ) = 0 
=>  x^2 = -1  or x =4
 
But  x^2  cannot be negative 
=> x = 4 is a solution
=> 1 real solution

So,  Answer will be B 
Hope it helps!

Watch the following video to MASTER Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

eaat · Answer

I found that x = 4; is there any way to confirm that this is the  only solution?

vasu1104 · Answer

f(x) = x^3 -4x^2+x-4

here from first two terms, we can take x^2 common.

x^2(x-4)+ x-4

now from entire equation (x-4) can be taken as common.

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

so, x-4 = 0
hence x =4

for, x^2+1 =0
x^2 = -1
this is not possible.

so only one value for x and that is x=4

option B

VividSloth · Answer

one real solution after factorizing the cubic polynomial we get (x^2+1) and (x-4) hence x = 4 is the only real solution

KarishmaB · Answer

The cubic will have at least 1 real solution. How many exactly there are 1/2/3, we can find once we find the first solution. The first solution is found by hit and trial. x = 1 or 2 are too small but x = 4 works.

f(x) = x^{3} - 4x^{2} + x - 4 = (x-4)(x^2 + 1) 
We obtain the quadratic by matching the coefficients on the LHS with RHS.  x^2  can never be negative so  x^2 + 1  can never be 0. So it has no real roots.

Hence the cubic has a single real root only.

Answer (B)

Note: An easier solution would be if you notice the symmetry of the coefficients to realize that 
 x^{3} - 4x^{2} + x - 4 
= x^{3} + x - 4x^{2} - 4 
= x(x^2 + 1) - 4(x^2 + 1) 
= (x - 4)(x^2 + 1)

If f(x) = x^3 - 4x^2 +x -4, then f(x) = 0 for how many real values of

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GMAT Problem Solving (PS) Questions

If f(x) = x^3 - 4x^2 +x -4, then f(x) = 0 for how many real values of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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Perfect 805 Score Debrief by Julia

My Rewards