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Paha777 [63]
2 years ago
10

Find a polynomial function of degree 3 with 2, i, -i as zeros.

Mathematics
1 answer:
Simora [160]2 years ago
5 0

Answer:

p(x)= x^3-2x^2+x-2

Step-by-step explanation:

Here we are given that a polynomial has zeros as 2 , i and -i . We need to find out the cubic polynomial . In general we know that if \alpha , \ \beta \ \& \ \gamma are the zeros of the cubic polynomial , then ,

\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma)

Here in place of the Greek letters , substitute 2,i and -i , we get ,

\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i)

Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,

\sf  \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\}

Simplify using i = √-1 ,

\sf \longrightarrow p(x)= (x-2)( x^2 + 1 )

Multiply by distribution ,

\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1)

Simplify by opening the brackets ,

\sf\longrightarrow p(x)= x^3+x-2x^2-2

Rearrange ,

\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}

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