✖️Complex roots

• Definition: Complex roots are solutions to polynomial equations that involve imaginary numbers, typically expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit. These roots often appear in conjugate pairs, meaning that if a + bi is a root, then a - bi is also a root.

Solving complex roots

• With the given equation, perform rational root theorem to identify potential rational roots, and then use synthetic division to simplify the polynomial, ultimately leading to the extraction of any complex roots present.

• Remmember that if f(x)=x^𝐧, then f(x) has 𝐧 roots

• All complex roots (including radicals and imaginaries) must come in pairs or conjugates, so if -2+𝐢 is a root, using the inverse according to the zero product property and the conjugates, the two zeros would be:

                            (x+2+𝐢)(x+2-𝐢)

• Then this must be triple-distributed to each other to get a polynomial, which you will long divide your original equation by to find other zeros.

Example:

1. Consider the polynomial equation: $f(x) = x^3 + 3x^2 + 2x + 4$

   • Apply the rational root theorem to find potential rational roots.

   • Use synthetic division to test for roots.

   • Suppose we find that $x = -1$ is a root.

   • Perform synthetic division:

     • $f(x) ÷ (x + 1)$ yields a quotient of $x^2 + 2x + 4$.

   • Now, find the roots of $x^2 + 2x + 4$ using the quadratic formula:

     • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

     • Here, $a = 1, b = 2, c = 4$:

     • The discriminant is $2^2 - 4(1)(4) = 4 - 16 = -12$ (negative, indicating complex roots).

     • Thus, roots are:

     • $x = \frac{-2 \pm \sqrt{-12}}{2} = -1 \pm i\sqrt{3}$.

Example 2:

2. Given the polynomial: $g(x) = x^2 + 4$

   • We can rewrite it as $g(x) = x^2 + 0x + 4$.

   • Using the quadratic formula to find roots:

     • $x = \frac{-0 \pm \sqrt{0^2 - 4(1)(4)}}{2(1)}$

     • Discriminant: $0^2 - 16 = -16$ (also negative).

     • Roots will be:

     • $x = \frac{0 \pm 4i}{2} = \pm 2i$ .