Write a polynomial equation that has three real zeroes, and a y intercept of 250

Forget about imaginary roots.

Draw an x axis and a y-axis, at right angles.

Place a dot at y=250.

Place a dot at x=60 and place a dot at x=120.

Sketch a graph: in quadrant II, go left from the y-intercept of 250, downward and then curve upward before getting to the x-axis. You have created a local min in the 2nd quadrant.

In the first quadrant, curve smoothly fromthe y intercept down to 60 on the x axis, and continue toward the right and into the fourth quadrant, eventually curving upward and eventually crossing the x axis at 120.  Continue upward indefinitely. You have created another local min, at about x=90.

The smooth polynomial curve will have a local max at(0,250).

Now y = a(x-60)(x-120)(x2+1) can represent this curve; it has two real zeros. Adjust "a" so that the y-intercept is 250.

Step-by-step explanation:

i hope this helps

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation:

the correct answer is the second option