Question

What is the polynomial function of lowest degree with lead coefficient 1 and roots i, -2 and 2?

Answer 1

Answer:

Step-by-step explanation:

the polynomial function is;  (x - i)(x-2)(x+2)

Answer 2

Answer:

$P(x)=x^3-ix^2-4x+4i$

Step-by-step explanation:

We have to find the polynomial of lowest degree with lead coefficient 1 and roots i, -2 and 2.

A polynomial can be written as:

$P(x)=a*(x-x_1)*(x-x_2)*...*(x-x_n)$

Where $a$ is the lead coefficient. And $x_1,x_2,...,x_n$ are the roots of the polynomial.

Then we have $a=1$ and,

$x_1=i\\x_2=-2\\x_3=2$

We can write the polynomial as:

$P(x)=1(x-i)(x-(-2))(x-2)\\P(x)=(x-i)(x+2)(x-2)$

You can apply squared binomial to $(x+2)(x-2)$:

$(x+2)(x-2)=x^2-2^2=(x^2-4)$

Then,

$P(x)=(x-i)(x^2-4)$ apply distributive property:

$P(x)=(x-i)(x^2-4)\\P(x)=x^3-4x-ix^2+4i\\P(x)=x^3-ix^2-4x+4i$

The the polynomial of lowest degree with leaf coefficient 1 and roots i, -2 and 2 is:

$P(x)=x^3-ix^2-4x+4i$