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

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.