Find possible values of the constant b

Question

the roots of the equation x² + bx + 24 = 0 are integers. Find the possible values of the constant b.

Answer 1

x² + bx + 24 = 0      looks like a polynomial that is product of multiplying two binomials.  The first term is x² and the last term is 24.  so if we write this using FOIL,

 

 

(x + 24)(x + 1) = 0

(x - 24)(x - 1) = 0

 

(x + 6)(x + 4) = 0

(x - 6)(x - 4) = 0

 

(x + 8)(x - 3) = 0

(x - 8)(x - 3) = 0

 

(x + 12)(x + 2) = 0

(x - 12)(x - 2) = 0

 

 

Now for each pair you see before you, find the middle term when multiplying the binomials. The coefficient of the middle term are your possible values of b. 

 

Answer 2

If the equation is written as Ax^2 + Bx + C = 0, then the product of the roots is C/A.

 

Now, using our knowledge of prime factors, we write 24 as 2*2*2*3.  With A=1, that means that we must divide 2*2*2*3 into the product of two integer roots.  Our choices are the sets:

   {1,24}, {-1,-24},

    {2,12}, (-2,-12},

     {4,6}, {-4,-6},

     {8,3}, {-8,-3}

 

The sums [the value of (-B/A)] are ±25, ±14, ±10, and ±11.  These are all the possible values for B  [since A=1].

Answer 3

If Ax²+Bx+C=0, then sum of roots = -B/A and product of roots = C/A

 

So, for the given equation, sum of roots = -b and product of roots = 24

 

Since the roots are required to be integers, the following are the possibilities:  1 and 24, -1 and -24, 2 and 12, -2 and -12, 3 and 8,

-3 and -8, 4 and 6, -4 and -6 

 

So, the possible values of b are: 25, -25, 14, -14, 11, -11, 10, -10