Question

Find all real and complex zeros of p(x)=x^3-x^2-4x-6​

Answer 1

Answer:

x = 3 , -1 + $\sqrt{-1}$ , -1 - $\sqrt{-1}$

Step-by-step explanation:

⇒ $p(x)=x^{3} -x^{2} -4x-6$

⇒ $p(x)=(x-3)(x^{2} +2x+2)$

now, for zeros of $(x^{2} +2x+2)$,

⇒ x =  $\frac{-b}{2a}$ ± $\frac{\sqrt{b^{2}-4ac } }{2a}$

⇒ x =  $\frac{-2}{2}$ ± $\frac{\sqrt{2^{2}-4(2) } }{2}$

⇒ x = -1 ± $\sqrt{-1}$

hence, all the roots are,

x = 3, -1 + $\sqrt{-1}$, -1 - $\sqrt{-1}$