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

Sagot :

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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]

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