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
6x^3y^9/36x^3y^-2 The x^3 values cancel each other out leaving the 1/6
(1/6)y^9/(y^-2) the y^-2 causes the y^9 to turn into y^11 (9+2=11)
(1/6)y^11/1 the 1/6 is multiplied by y^11
y^11/6
Between 1 and 2 would be decimals.
E.g: 1.1, 1.2,1.3 etc
<em>Please refer to the attachment for the answer and explanation. Hope it helps!!</em>
2 because you can evenly split it in half 2 different ways
