Respuesta :
We are given roots of a polynomial function : i, –2, and 2.
And leading coefficient 1 .
We need to find the polynomial function of lowest degree.
Please note: We have one root i, that is a radical root. And a radical always comes in pair of plug and minus sign.
Therefore, there would be another root -i.
So, we got all roots of the polynomial function : i, -i, -2, and 2.
For the given roots, we would have factors of the polynomial (x-i)(x+i)(x+2)(x-2).
Now, we need to multiply those factors to get the polynomial function.
[tex]\mathrm{Expand}\:\left(x-i\right)\left(x+i\right):\quad x^2+1[/tex]
[tex]\left(x+2\right)\left(x-2\right):\quad x^2-4[/tex]
[tex]\left(x-i\right)\left(x+i\right)\left(x+2\right)\left(x-2\right)=\left(x^2+1\right)\left(x^2-4\right)[/tex]
[tex]\mathrm{Expand}\:\left(x^2+1\right)\left(x^2-4\right)=x^4-4x^2+ \:x^2-\:4[/tex]
[tex]=x^4-3x^2-4[/tex]
Therefore, correct option is 2nd option [tex]f(x)=x^4-3x^2-4[/tex].
Answer:
Option 2 is correct.
Step-by-step explanation:
Given the roots of the polynomial function. we have to find the lowest degree polynomial with leading coefficient 1 and roots i, –2, and 2.
By complex conjugate root theorem which states that if P is the polynomial and a+ib is a root of P with a and b real numbers, then its complex conjugate a-ib is also a root of that polynomial P.
∴ -i is also the root of the polynomial function.
Hence, there are 4 roots of the given polynomial function f(x)
f(x) can be written as
[tex]f(x)=(x+i)(x-i)(x+2)(x-2)[/tex]
[tex]=(x^2-i^2)(x^2-2^2)[/tex]
[tex]=(x^2+1)(x^2-4)[/tex]
[tex]=x^4-4x^2+x^2-4[/tex]
[tex]=x^4-3x^2-4[/tex]
Option 2 is correct.
