How do you find all the real and complex roots of x^3 - x^2 - 7x + 15=0x3−x2−7x+15=0?

1 Answer
George C.
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 = -3x=−3, x = 2+-ix=2±i

Explanation:

f(x) = x^3-x^2-7x+15f(x)=x3−x2−7x+15

By the rational root theorem, any rational roots of f(x) = 0f(x)=0 must be expressible in the form p/qpq for some integers pp and qq where pp is a factor of the constant term 1515 and qq is a factor of the coefficient 11 of the leading term.

That means that the only possible rational roots are:

+-1±1, +-3±3, +-5±5, +-15±15

Let's try the first few...

f(1) = 1-1-7+15 = 8f(1)=1−1−7+15=8

f(-1) = -1-1+7+15 = 20f(−1)=−1−1+7+15=20

f(3) = 27-9-21+15 = 12f(3)=27−9−21+15=12

f(-3) = -27-9+21+15 = 0f(−3)=−27−9+21+15=0

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

x^3-x^2-7x+15 = (x+3)(x^2-4x+5)x3−x2−7x+15=(x+3)(x2−4x+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)x=−b±√b2−4ac2a

= (4+-sqrt((-4)^2-(4*1*5)))/(2*1)=4±√(−4)2−(4⋅1⋅5)2⋅1

=(4+-sqrt(-4))/2 = 2+-i=4±√−42=2±i

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