How do you find all the real and complex roots of x^3 - x^2 - 7x + 15=0?

1 Answer
Jan 29, 2016

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:

f(x) = x^3-x^2-7x+15

By the rational root theorem, any rational roots of f(x) = 0 must be expressible in the form p/q for some integers p and q where p is a factor of the constant term 15 and q is a factor of the coefficient 1 of the leading term.

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 is a root of f(x) = 0 and (x+3) is a factor of f(x)

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