What is the product of all possible values of x for the equation

Question

What is the product of all possible values of x for the equation  4x^3 -23x^2 + 26x +8 = 0 ?

A. -4
B. -2
C. 1
D. 2
E. 4

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TahaTT · Accepted Answer

Instead of the long calculation, could we simply use the formulas for the sum and product of roots of a polynomial?

Sum of roots = -b/a
Product of roots = z/a (if the degree of polynomial is even)
Product of roots = -z/a (if the degree of polynomial is odd)

Since this is a 3rd degree polynomial,
Product of roots = -z/a = -8/4 = -2
So the correct option is B.

sujoykrdatta · Answer

Let the roots of the equation be p, q, r and s
Thus, we can write:

4x^3 -23x^2 + 26x +8 = 4(x - p)(x - q)(x - r)(x - s)

Thus, comparing the constant term on both sides, we get:

8 = 4(-p)(-q)(-r)(-s)

=> 8 = 4pqrs

=> pqrs = 2

Thus, the product of the roots = 2

Answer D

randarth · Answer

It is a cubic equation. Only three roots possible.

Let the roots of the equation be p, q, and r
Thus, we can write:

4x^3 -23x^2 + 26x +8 = 4(x - p)(x - q)(x - r)

Thus, comparing the constant term on both sides, we get:

8 = 4(-p)(-q)(-r)

=> 8 = -4pqr

=> pqr = -2

Thus, the product of the roots = -2

GMATGuruNY · Answer

Since the last term is 8, one of the roots is probably a factor of 8: 1, 2, 4, 8 When we test x=2, we get: 4x^3 -23x^2 + 26x + 8 = (4*2^3) - (23*2^2) + (26*2) + 8 = 32 - 92 + 52 + 8 = 0 Implication: (x-2) is a factor of 4x^3 -23x^2 + 26x + 8 Dividing x-2 into 4x^3 -23x^2 + 26x + 8 , we get: poly 4.png Factoring 4x^2 - 15x - 4 = 0 , we get: (4x+1)(x-4) = 0 Thus: 4x^3 -23x^2 + 26x + 8 = (x-2)(4x+1)(x-4) Roots = 2, -1/4, 4 Product of the roots = = 2*-\frac{1}{4}*4 = -2 B

Anvesh99 · Answer

Hello  sujoykrdatta ,

Can you please help me understand the question as well as your approach for this question. I am unable to understand.

Thanks

NirupaD · Answer

General Polynomial
 f(x) = ax^n + bx^(n-1) + cx^(n-2) + ... + z

And then factor it like this:

f(x) = a(x−p)(x−q)(x−r)...

Then p, q, r, etc are the roots (where the polynomial equals zero)

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)

Cubic
Now let us look at a Cubic (one degree higher than Quadratic):

ax^3 + bx^2 + cx + d

As with the Quadratic, let us expand the factors:

a(x−p)(x−q)(x−r)
=  ax^3 − a(p+q+r)x^2 + a(pq+pr+qr)x − a(pqr)

And we get:

Cubic:  ax^3 +bx^2 +cx +d 
Expanded Factors:  a^3 −a(p+q+r)x^2 +a(pq+pr+qr)x −apqr 
We can now see that  −a(p+q+r)x^2 = bx^2,  so:

−a(p+q+r) = b
p+q+r = −b/a
And −apqr = d, so:

pqr = −d/a

Adding the roots gives −b/a (exactly the same as the Quadratic)
Multiplying the roots gives −d/a (similar to +c/a for the Quadratic)
(We also get pq+pr+qr = c/a, which can itself be useful.)

Higher Polynomials
The same pattern continues with higher polynomials.

In General:

Adding the roots gives −b/a
Multiplying the roots gives (where "z" is the constant at the end):
z/a (for even degree polynomials like quadratics)
−z/a (for odd degree polynomials like cubics)

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What is the product of all possible values of x for the equation

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