If 6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0, what is the product of all poss

Question

If  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 , what is the product of all possible value of x ?

A. -5/6
B. -1/6
C. 1/6
D. 5/6
E. 1

6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0

Find possible values of x.

yashikaaggarwal · Accepted Answer

If we have a general polynomial like this:
f(x) = axn + bxn-1 + cxn-2 + ... + z

Then:
Adding the roots gives −b/a
Multiplying the roots gives:
=> z/a (for even degree polynomials like quadratics)
=> −z/a (for odd degree polynomials like cubics)

6x^4-25x^3+12x^2+25x+6=0
value of x = roots of equation
Here degree of equation is 4, therefore the product of all values of roots of x = 6/6= 1

IMO E

RaghavKhanna · Answer

For a 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)

=> Product of roots of 6x^4−25x^3+12x^2+25x+6=0 will be: (-1)^4 * (6/6) =  1

=>  Answer: E

rajatchopra1994 · Answer

Explanation:

6x^4 − 25x^3 + 12x^2 + 25x + 6 = 0
By trial error method, Put x=-1,1,2,3 to find atleast 2 roots of above equation:
So Putting x=2 
= 6(2)^4 − 25(2)^3 + 12(2)^2 + 25(2) + 6
= 6×16−25×8+12×4+75+6
= 96−200+48+56
= −200 + 200
=0
(x−2) is a one factor.
Now put x=3 in Equation.
= 6(3)^4 − 25(3)^3 + 12(3)^2 + 25(3) + 6
= 6×81−25×27+12×9+25×3+6
= 486−675+108+75+6
= −675+675
=0
(x−3) is an also a factor.

Now (x−3)(x−2)=x^2 − 5x + 6 is a factor of Eqaution.

(x^2 −5x+6) | 6x^4 − 25x^3 + 12x^2 + 25x + 6 | ( 6x^2 + 5x +1 )
 6x^4 − 30x^3 + 36x^2
 ______________________
 5x^3 − 24x^2 + 25x + 6
 5x^3 − 25x^2 + 30x
 _____________________
 x^2 − 5x + 6
 x^2 − 5x + 6
 _______________
 0
6x^4 − 25x^3 +12x^2 +25x+6=0 can be written as below:
(x−2)(x−3)(6x^2 + 5x + 1)=0
(x−2)(x−3)(6x^2 + 3x + 2x +1)=0
(x−2)(x−3)(3x*(2x+1)+1*(2x+1))=0
(x−2)(x−3)(3x+1)(2x+1)=0
x = 2, 3, −1/3, −1/2.

Product of all roots = 1

IMO-E

GMATinsight · Answer

CONCEPT : 
Any equation of the form  ax^4 + bx^3 + cx^2 + dx+ e = 0  has four roots p, q, r, s

Sum of the roots,  ∑p = p+q+r+s = \frac{-b}{a} 
Sum of the roots two at a time,  ∑pq = pq+pr+ps+qr+qs+rs = \frac{c}{a} 
Sum of the roots three at a time,  ∑pqr = pqr+pqs++prs+qrs = \frac{-d}{a} 
Product of the roots,  ∑pqrs = pqrs = \frac{e}{a}

Comparing  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0  with  ax^4 + bx^3 + cx^2 + dx+ e = 0 
a = 6, b = -25, c = 12, d = 25 and e = 6

Product of roots of  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0  is given by  = \frac{e}{a} = \frac{ 6}{6} = 1

Answer: Option E

Kinshook · Answer

Asked: If  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 , what is the product of all possible value of x ?

Product of all possible values (roots) of x = 6/6 = 1

It is assumed that all possible values means roots of the equation.

IMO E

Kinshook · Answer

Hi  GMATinsight 
 Product of all possible values of x   may be different than  product of all roots  for the equation in case one or more roots are repeated.
Incidentally, in this equation all roots are different and both answers are same.

What is your view on this for all type of questions ?

ScottTargetTestPrep · Answer

Solution:
 
The product of all possible roots of a polynomial equation is given by the following formula:

(-1)^n * k/a

where n is the degree of the polynomial, k is the constant term and a is the leading coefficient.

Therefore, the product of all possible roots of the given polynomial equation is:

(-1)^4 * 6/6 = 1

Answer: E

hiranmay · Answer

If  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0 , what is the product of all possible value of x ?

A. -5/6
B. -1/6
C. 1/6
D. 5/6
E. 1 -->   correct :  6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0  =>  x^4 - (25/6)x^3 + 2x^2 + (25/6)x + 1 = 0 , diving by 6 => let's say the equation has 4 values of x: a, b, c & d, so the equation is (x-a)(x-b)(x-c)(x-d)=0 => x^4 + ..x^3+..x^2+..x+acbd =0 => so the product of all possible value of x = 1

BlueRocketAsh · Answer

In ax^4 + bx^3 + cx^2 +dx + e = 0 type of equation, product of the roots is given by e/a
Therefore answer is 6/6=1

bumpbot · Answer

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