Find the sum and the product of the roots (real and complex) of x^3 + 3x^2 + 7x − 11 = 0
x3+3x2+7x−11=0
simplifies to:
(x−1)(x2+4x+11)=0
and
x2+4x+11=0
which gives an imaginary number
but for the complex part,
you get the roots of x as:
x=−4±√42−4∗1∗112
whic becomes
x=2±√−7
so the roots of
x3+3x2+7x−11=0
are
x=1
and
x=2±√−7
In response to your question:
A good way to do this is by the rational root theorem.
For any polynomial, anxn+an−1xn−1…a1x1+a0, the rational roots can all be expressed as:±factor of a0factor of an.
Therefore in this polynomial, the only possible rational roots are ±1,±11. (factors of a0 are 1, 11 and the only factor of an is 1)
Plug these values into the function, to see which ones produce zero. ±1 is a really, really common ones, so always try those first.
Remainder theorem is always helpful. Remember if for a function f(x), f(a)=0, then x−a is a factor of f(x).
Here is the workthrough of the first step:
Use rational root theorem, plug in 1 to f(x)=x3+3x2+7x−11.
13+3∗12+7∗1−11=0.
Therefore we know because f(1)=0, then x-1 is a factor of f(x), so we can factor it out.