MATH SOLVE

5 months ago

Q:
# What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?f(x) = x3 – x2 – 4x + 4f(x) = x4 – 3x2 – 4f(x) = x4 + 3x2 – 4f(x) = x3 + x2 – 4x – 4

Accepted Solution

A:

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].