Question

6. Consider the polynomial p(x) = x^2 + kr^2 +x+6. Find a value of k so that x+1 is a factor of P. Find all the zeros of P.​

Answer 1

If x + 1 is a factor of p(x) = x³ + k x² + x + 6, then by the remainder theorem, we have

p (-1) = (-1)³ + k (-1)² + (-1) + 6 = 0   →   k = -4

So we have

p(x) = x³ - 4x² + x + 6

Dividing p(x) by x + 1 (using whatever method you prefer) gives

p(x) / (x + 1) = x² - 5x + 6

Synthetic division, for instance, might go like this:

-1   |   1    -4     1     6

...   |         -1    5    -6

----------------------------

...   |   1    -5    6     0

Next, we have

x² - 5x + 6 = (x - 3) (x - 2)

so that, in addition to x = -1, the other two zeros of p(x) are x = 3 and x = 2