Form a polynomial whose zeros and degree are given. ​Zeros: −3​, 3​, 2​; ​ degree: 3 Type a polynomial with integer coefficients and a leading coefficient of 1

If a polynomial P(x) has a zero equal to a, then (x-a) is a factor of this polynomial. So if a polynomial has zeros a, b and c then it has we could write:

         P(x)=(x-a)(x-b)(x-c).

Here we can clearly see that a, making the left hand side 0 because of the factor (x-a), makes the left hand side 0 as well. This means that P(a)=0. This illustrates the discussion above.


Thus, substituting a, b, c with −3​, 3​, 2 we can write P(x)=(x+3)(x-3)(x-2).

We can expand the right hand side to have the polynomial in standard form:

[tex]P(x)=(x+3)(x-3)(x-2)=P(x)=[(x+3)(x-3)](x-2)[/tex] [tex]=(x^2-9)(x-2)=x^3-2x^2-9x+18[/tex]

We see that all conditions are satisfied.



Answer: [tex]P(x)=x^3-2x^2-9x+18[/tex]