How do you find all the real and complex roots of x^3 - x^2 - 7x + 15=0?
1 Answer
Use the rational root theorem to help find one root, then factor that out to find the remaining two Complex roots:
x = -3 ,x = 2+-i
Explanation:
By the rational root theorem, any rational roots of
That means that the only possible rational roots are:
+-1 ,+-3 ,+-5 ,+-15
Let's try the first few...
f(1) = 1-1-7+15 = 8
f(-1) = -1-1+7+15 = 20
f(3) = 27-9-21+15 = 12
f(-3) = -27-9+21+15 = 0
So
x^3-x^2-7x+15 = (x+3)(x^2-4x+5)
The remaining quadratic factor has negative discriminant, so no Real roots, but we can find its Complex roots using the quadratic formula:
x = (-b+-sqrt(b^2-4ac))/(2a)
= (4+-sqrt((-4)^2-(4*1*5)))/(2*1)
=(4+-sqrt(-4))/2 = 2+-i