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zloy xaker [14]
3 years ago
12

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

Mathematics
1 answer:
yuradex [85]3 years ago
7 0

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}

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