Find a polynomial function of least degree having only real coefficients, a leading of 1, and zeros of 2 and 2+i.

y = x³ − 6x² + 13x − 10

Step-by-step explanation:

Complex roots come in pairs, so if 2 + i is a root, then 2 − i is also a root.

y = (x − 2) (x − (2 + i)) (x − (2 − i))

y = (x − 2) (x − 2 − i) (x − 2 + i)

y = (x − 2) ((x − 2)² − i²)

y = (x − 2) (x² − 4x + 5)

y = x (x² − 4x + 5) − 2 (x² − 4x + 5)

y = x³ − 4x² + 5x − 2x² + 8x − 10

y = x³ − 6x² + 13x − 10

definition: a circle is the  set of all points in a plane that are equidistant  from a given  point  called the center of the circle. we use the symbol ⊙ to represent a circle. the a line segment from the center of the circle to  any point  on the circle is a radius of the circle.

step-by-step explanation:

i googled it and this is what i got i don't know how it is. sorry.

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.