# The polynomial p(x) = x^3 - x + 6. p(x)=0 has one root equal to -2. Show that the equation has no other real roots.

*print*Print*list*Cite

### 1 Answer

The polynomial p(x) = x^3 - x + 6 = 0 has one root equal to -2

x^3 - x + 6 = 0

=> x^3 + 2x^2 - 2x^2 - 4x + 3x + 6 = 0

=> x^2(x + 2) - 2x(x + 2) + 3(x + 2) = 0

=> (x^2 - 2x + 3)(x + 2) = 0

It can be seen that in the quadratic equation x^2 - 2x + 3, 2^2 - 4*1*3 = 4 - 12 = -8. As b^2 - 4ac < 0 the quadratic equation only has complex roots.

**This shows that p(x) = 0 has only one real root x = -2**