Polynomial Division Questions

Question

The polynomial
$$g(x) = x^3 - x^2 - (m^2 + m) x + 2m^2 + 4m + 2$$
is divisible by $x-4$ and all of its zeroes are integers. Find all possible values of $m$.

Answer 1

g(x) = x³ - x² - (m² + m)x + 2m²+ 4m + 2

 

Since g(x) is divisible by x-4, g(4) = 0.

 

So, 64 - 16 - 4(m² + m) + 2m²+ 4m + 2 = 0

 

Simplify to obtain:  -2m²  + 50 = 0

 

                                         m²= 25

 

                                         m = ±5

 

If m = 5, then g(x) = x³ - x²  - 30x + 72 

 

                             = (x-4)(x+6)(x-3)

 

     The roots are all integers  (4, -6, and 3)

 

If m = -5, then g(x) = x³ - x² - 20x + 32

 

                              = (x-4)(x² + 3x - 8)

 

      The roots aren't all integers.

 

The only value of m that works is 5.  

 

 

Answer 2

We can use synthetic division.  Divide the coefficients of g(x) by the root.  The remainder my be zero in order for g(x) to be division by (x - 4).

 

 

4 |  1       -1            (-m² - m)              (2m² + 4m + 2)

 

                4                 12                    (-4m² - 4m + 48)

_______________________________________________

      1        3           (-m² - m + 12)                    0

 

 

Solve this equation for m.  From the last column in the division process.

 

2m² + 4m + 2 - 4m² - 4m + 48 = 0

 

-2m² + 50 = 0