desiree2021 desiree2021
  • 02-11-2019
  • Mathematics
contestada

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

Respuesta :

arindamvutla arindamvutla
  • 07-11-2019

Answer:

x = 3 , -1 + [tex]\sqrt{-1}[/tex] , -1 - [tex]\sqrt{-1}[/tex]

Step-by-step explanation:

⇒ [tex]p(x)=x^{3} -x^{2} -4x-6[/tex]

⇒ [tex]p(x)=(x-3)(x^{2} +2x+2)[/tex]

now, for zeros of [tex](x^{2} +2x+2)[/tex],

⇒ x =  [tex]\frac{-b}{2a}[/tex] ± [tex]\frac{\sqrt{b^{2}-4ac } }{2a}[/tex]

⇒ x =  [tex]\frac{-2}{2}[/tex] ± [tex]\frac{\sqrt{2^{2}-4(2) } }{2}[/tex]

⇒ x = -1 ± [tex]\sqrt{-1}[/tex]

hence, all the roots are,

x = 3, -1 + [tex]\sqrt{-1}[/tex], -1 - [tex]\sqrt{-1}[/tex]

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