+11 Find the sum and the product of the roots (real and complex) of x^3 + 3x^2 + 7x − 11 = 0
+60 \(x^3 + 3x^2 + 7x − 11 = 0\)
simplifies to:
\((x-1)(x^2+4x+11)=0\)
and
\(x^2+4x+11=0\)
which gives an imaginary number
but for the complex part,
you get the roots of x as:
\(x=\frac{-4±\sqrt{4^2-4*1*11}}{2}\)
whic becomes
\(x=2±\sqrt{-7}\)
so the roots of
\(x^3 + 3x^2 + 7x − 11 = 0\)
are
\(x = 1\)
and
\(x=2±\sqrt{-7}\)
+399 In response to your question:
A good way to do this is by the rational root theorem.
For any polynomial, \(a_nx^n+a_{n-1}x^{n-1}\dots a_1x^1+a_0\), the rational roots can all be expressed as:\(\pm\frac{\text{factor of }a_0}{\text{factor of }a_n}\).
Therefore in this polynomial, the only possible rational roots are \(\pm 1, \pm11\). (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. \(\pm1\) 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 \(f(x)={x}^{3}+3x^{2}+7x-11\).
\({1}^{3}+3*{1}^{2}+7*1-11=0\).
Therefore we know because \(f(1) = 0\), then x-1 is a factor of \(f(x)\), so we can factor it out.
