Answer:
[tex] p(x)= x^3-2x^2+x-2 [/tex]
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 [tex]\alpha , \ \beta \ \& \ \gamma [/tex] are the zeros of the cubic polynomial , then ,
[tex]\sf \longrightarrow p(x)= (x -\alpha )(x-\beta)(x-\gamma) [/tex]
Here in place of the Greek letters , substitute 2,i and -i , we get ,
[tex]\sf\longrightarrow p(x)= (x -2 )(x-i)(x+i) [/tex]
Now multiply (x-i) and (x+i ) using the identity (a+b)(a-b)=a² - b² , we have ,
[tex]\sf \longrightarrow p(x)= (x-2)\{ x^2 - (i)^2\} [/tex]
Simplify using i = √-1 ,
[tex]\sf \longrightarrow p(x)= (x-2)( x^2 + 1 ) [/tex]
Multiply by distribution ,
[tex]\sf \longrightarrow p(x)= x(x^2+1) -2(x^2+1) [/tex]
Simplify by opening the brackets ,
[tex]\sf\longrightarrow p(x)= x^3+x-2x^2-2 [/tex]
Rearrange ,
[tex]\sf\longrightarrow \underline{\boxed{\blue{\sf p(x)= x^3-2x^2+x-2}}}[/tex]
