How do you find a polynomial of degree 2 that has a zero of -2?

1 Answer
Nov 15, 2016

P(x) = x^2+(k-2)x-2kP(x)=x2+(k2)x2k where kk is a real number.

Explanation:

A polynomial P(x)P(x) has some number alphaα as a zero if and only if x-alphaxα is a factor of P(x)P(x). To generate a polynomial with desired zeros, then, we can multiply any such factors.

As our desired polynomial has -22 as a zero, it must have a factor of x-(-2) = x+2x(2)=x+2. As no other specific zero is given, we can make that choice ourselves. Suppose the other zero (possibly also being -22), is kk. Then the polynomial would be

P(x) = (x+2)(x-k)P(x)=(x+2)(xk)

=x^2+(2-k)x - 2k=x2+(2k)x2k

Choosing any value for kk will give a degree 22 polynomial with -22 as a zero. For example, k=0k=0 gives x^2+2xx2+2x, or k=2k=2 gives x^2-4x24. Multiplying by any nonzero constant also will result in a valid polynomial.