What is the product of all possible values of x for the equation x^3 -

Question

What is the product of all possible values of x for the equation x^3 - 5x^2 + 8x - 4= 0?

A. 4
B. 2
C. -4
D. -2
E. 8

Source: GMAT Quant

rajatchopra1994 · Answer

Explanation:

IF the question say, a product of all roots then below is the solution:

For Polynomial: a0x^n + a1x^(n-1) + a2x^(n-2) + .... + an = 0
The sum of product of roots taken 'n' at a time = (-1)^n * (an/a0): (-1)^3 * (-4/1) = 4

IMO- A   B

x^3−5x^2+8x−4=0
(x−1)*(x−2)^2=0
x=1,2,2
In this case, we have two roots which are similar i.e. x=2, 
Possible values are 1 and 2.
Product of all possible values =1x2=2.

IMO-B

rcgard · Answer

x^3-5x^2+8x-4 =0
 x(x^2-5x+8)-4 =0
 x(x^2-5x+4+4)-4 =0
 x((x-1)(x-4)+4)-4 =0
 (x^2-4x)(x-1)+4x-4 =0
 x^2-4x)(x-1)+4(x-1) =0
 (x-1)(x^2-4x+4) =0
(x-1)(x-2)(x-2)=0
x=1 or x=2
1*2 = 2
Answer choice B IMO

saurabh851 · Answer

Ans: B (2)

x^3-5x^2+8x-4 = 0
x(x^2-5x+8) = 4

Case 1: x=1 and x^2-5x+8 = 4
on solving, x = 1,4

Case 2: x=2 and x^2-5x+8 = 2
on solving, x=2,3

Four values of x = 1,2,3,4
But only 1,2 satisfying main equation

Product = 2

AnirudhaS · Answer

**Solution Edited. Methods 1 and 2 are for finding the product of the roots. In this case we have two roots which are similar x=2, so possible values are 1 and 2.
Product of all possible values =1*2=2

Answer:  B

**I am still keeping the two methods of solving product of roots in, if it helps someone BUT they cannot be used in this particular problem.

Method 1: 
If we have a general polynomial like this:
 f(x) = ax^n + bx^{n-1} + cx^{n-2} + ... + z 
Then Product of roots 
=  \frac{z}{a}  (for even degree polynomials like quadratics)
=  −\frac{z}{a}  (for odd degree polynomials like cubics)
Product of roots =  -\frac{(-4)}{1}=4

Method 2: 
 x^3-5x^2+8x-4 
= (x-1)(x-2)^2 
Product of roots = 1*2*2=4

Papist · Answer

x=2 is one solution, so it shouldn't it be 1*2, not 1*2*2?

AnirudhaS · Answer

you might be right, it does say "possible values" and not "all roots". I'll wait for the OA and amend (or not amend) the answer accordingly.

IanStewart · Answer

I have no idea what that source is, but you don't need to factor proper cubics on the GMAT, at least when grouping is impossible, and you don't need to know any theorems about their roots. So this question is well out of scope.

If a cubic begins with "x^3" (and not something more complicated like "5x^3") then the standard way to factor, if you can't group terms, is to focus on the number at the end, "-4". If we can factor this cubic into something like (x - a)(x - b)(x - c) (which isn't always possible), then the product -abc will be equal to -4. So a, b and c will be factors of 4. So we can just plug the factors of 4, including negative ones, into the original equation until we find a solution that works. This can take a long time. Here if you plug in x = 1 first, you get lucky on your first try, but if instead you try x = 4 and x = -4 first, for example, the question already takes an inordinate amount of time. The GMAT is not a test of how luckily you guess when you plug numbers into things, so that's already a reason these types of questions are unrealistic.

If x = 1 is a solution to the cubic, then x-1 is a factor of the cubic. So we know

(x - 1)(x^2 + bx + c) = x^3 - 5x^2 + 8x - 4

c must equal 4 to give us the product of -4 at the end, so

(x - 1)(x^2 + bx + 4) = x^3 - 5x^2 + 8x - 4

and if we multiply out the left side, we find we get a (b -1)x^2 term (you could instead compare the x terms if you wanted to). Since we have -5x^2 on the right, b - 1 = -5, and b = -4. So we have just shown that

x^3 - 5x^2 + 8x - 4 = (x - 1)(x^2 - 4x + 4)

and we can finish factoring:

x^3 - 5x^2 + 8x - 4 = (x - 1)(x^2 - 4x + 4) = (x - 1)(x - 2)^2

Since this cubic equals zero, it has two solutions, x = 1 and x = 2. These are the only two possible values of x, and their product is 2, so that is the answer.

You'll never need to solve a problem like this on the GMAT, but you might need some of the math above in a much simpler context, so there might be a reason to understand it at least.

The lone situation where I've needed to factor a genuine four-term cubic on an official question is one where a 'grouping' technique is available, and even then, the question usually completes the first step for you. If you saw a cubic like this one:

x^3 + 5x^2 - 9x - 45 = 0

you can group the terms in pairs and lift out a common factor:

x^2(x + 5) - 9(x + 5) = 0

and now (x + 5) is a common factor in both terms, so we can factor it out:

(x^2 - 9)(x + 5) = 0
(x + 3)(x - 3)(x + 5) = 0

That's something I could imagine seeing on the GMAT, because you don't need any special knowledge about cubic polynomials to solve -- there are definitely questions at least where you need to see that you can factor (x+5) from x^2(x+5) - 9(x + 5).

mSKR · Answer

on seeing the answer choices , x can be 1,-1; 2 ,-2 or 4,-4

if x= -1 then equation is not zero
if x= 1 then euqation is zero
if x=2: the equation is 0 
if x= -2: the equation is not zero
if x= 4: the equation is not zero
if x= -4;the equatio is not zero

hence possible answer is x= 2 and x= 1 
multiply both factors : result = 2

hence B

AnirudhaS · Answer

I am no expert, but IMO this is not the right approach. 
Why did you exclude 8 or -8?
Also why did you assume the roots are integral values? Why cant the roots be something like 1/2 and 2, then product is 1? The possibilities are endless.

IanStewart · Answer

There's a theorem called the Integral Roots Theorem (a special case of the Rational Roots Theorem) which says that if a polynomial contains only integers (like the one in this question), and the coefficient of the term with the highest exponent is '1' (as we see here - we just have x^3, and not, say, 10x^3), then *if* the polynomial has any rational roots (which is not guaranteed), those roots will be positive or negative integer factors of the number at the end. So if the cubic in this question has any integer solutions, they will be positive or negative factors of 4.

So if you were going to guess (and it would just be a guess) that all of the solutions here were integers, the only candidates you'd need to check are 1, 2, 4, -1, -2 and -4.

I didn't do that, because there's really no way to know that the solutions will all be integers. The one thing we can be sure of is that the cubic has at least one solution, by the Intermediate Value Theorem from calculus: because the cubic is continuous and goes to -∞ when x is very negative, and goes to +∞ when x is very big, the cubic must at some point equal every value between -∞ and +∞. So it has to equal zero for some value of x. But that only guarantees it has one solution (which might not be an integer), which is why in my solution I just guessed there was one integer solution and used factoring to complete the problem.

Notice that I've mentioned above two mathematical facts (Rational Roots Theorem, Intermediate Value Theorem) you need no knowledge of whatsoever for the actual GMAT. That's why this question is miles beyond the scope of the test. In general, I strongly recommend working from sources that understand what the GMAT does and does not test, because otherwise you can waste a lot of time learning things that will be irrelevant to your GMAT success.

What is the product of all possible values of x for the equation x^3 -

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