Question

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

Answer 1

x = 3 , -1 +
`√(-1)` , -1 -
`√(-1)`

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 =
`(-b)/(2a)` ±
`\frac{\sqrt{b^(2)-4ac } }{2a}`

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

⇒ x = -1 ±
`√(-1)`

hence, all the roots are,

x = 3, -1 +
`√(-1)`, -1 -
`√(-1)`