Find a third-degree polynomial equation with rational coefficients that has roots –4 and 2 + i

1 answer:

8 0

If it has rational coefients and is a polygon
if a+bi is a root then a-bi is also a root

the roots are -4 and 2+i
so then 2-i must also be a root

if the rots of a poly are r1 and r2 then the factors are
f(x)=(x-r1)(x-r2)

roots are -4 and 2+i and 2-i

f(x)=(x-(-4))(x-(2+i))(x-(2-i))
f(x)=(x+4)(x-2-i)(x-2+i)
expand
f(x)=x³-11x+20

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Which is a solution for 3 > f? f = 7 f = 9 f= 3 f =2?

Fed [463]

Answer:

F = 2

Step-by-step explanation: If you want to know which one is a solution, you have to substitute the numbers in for F.

F = 7:

3 > f

3 > 7

This is false because 3 is NOT greater than 7, it is less than 7.

F = 9:

3 > f

3 > 9

This is false because 3 is NOT greater than 9, it is less than 9.

F = 2:

3 > f

3 > 2

This is TRUE because 3 IS greater than 2.

Hope this help you!!! :)

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