Find a second-degree polynomial (that is, an equation of the form y =a + bx + cx2 d that goes through the three points (0, 1), (1, 0), and (-1, 0). is there more than one possibility?

Step-by-step explanation:

we're given three points (0,1),(1,0) and (-1,0). and and equation of a parabola

we can plug in each of the coordinates, and 3 equations

(x,y) = (0,1)

(x,y) = (1,0)

(x,y) = (-1,0)

These are are three equations, well we can simultaneously solve them to find the values of a, b and c.

we already found that, a = 1. so we plug this value in the rest of the equations. we'll use equation B.

we can substitute this value of b and a = 1, equation C

we can use this value of c back in b

hence our equation of the 2-degree polynomial will be:

and this polynomial indeed passes through all the points (0, 1), (1, 0), and (-1, 0).

And since these are the only solutions to the simultaneous equation we solved (i.e. we have single values of a,b and c). there's no other possibility.

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.