Find a polynomial with integer coefficients that satisfies the given conditions.

P has degree 2 and zeros

2 + i and 2 − i.

clearly, one such

P(x) = (x-(2+i))(x-(2-i))
= ((x-2)-i)((x-2)+i)
= (x-2)^2 - i^2
= x^2-4x+4 + 1
= x^2-4x+5

extra credit: why does the question imply that there is more than one such P(x)?

Well, Polynomial P is feeling quite fortunate to have such complex friends as its zeros! And since complex numbers often come in pairs, we can say that the zeros are 2 + i and 2 - i. Now, when we have complex zeros, their conjugates are also zeros of the polynomial. So, the conjugates of 2 + i and 2 - i are 2 - i and 2 + i, respectively.

Now, to find the polynomial P, we can start by using the fact that the product of the zeros of a quadratic polynomial is equal to the constant term divided by the leading coefficient. Since P(x) has integer coefficients, we know that P(x) can be written in the form P(x) = a(x - z1)(x - z2), where z1 and z2 are the zeros of P(x).

Now let's plug in the given zeros into this form:

P(x) = a(x - (2 + i))(x - (2 - i))
= a(x - 2 - i)(x - 2 + i)
= a[(x - 2) - i][(x - 2) + i]
= a[(x - 2)^2 - i^2]
= a[(x - 2)^2 + 1]

So, our polynomial, P(x), is P(x) = a(x - 2)^2 + a. Since we want P to have integer coefficients, we can let a = 1 to get P(x) = (x - 2)^2 + 1.

So there you have it! The polynomial with integer coefficients that satisfies the given conditions is P(x) = (x - 2)^2 + 1. It's like P is saying, "Hey, I may have some complex roots, but I'm still an integer-coefficient polynomial!"

To find a polynomial with integer coefficients that satisfies the given conditions, we can use the fact that complex zeros always come in conjugate pairs.

Let's start by writing the factors in factored form using the given zeros:

(x - (2 + i))(x - (2 - i))

Now we can simplify this expression:

(x - 2 - i)(x - 2 + i)

Using the distributive property, we can expand this expression:

(x - 2 - i)(x - 2 + i) = (x - 2)^2 - (i)^2

Simplifying further,

(x - 2 - i)(x - 2 + i) = (x - 2)^2 - i^2
= (x - 2)^2 - (-1)
= (x - 2)^2 + 1

Therefore, the polynomial with integer coefficients that satisfies the given conditions is:

P(x) = (x - 2)^2 + 1

To find a polynomial with integer coefficients that satisfies the given conditions, we can use the fact that if a polynomial has a complex zero, then its conjugate is also a root.

The zeros given are 2 + i and 2 - i, which means their conjugates are also zeros: 2 - i and 2 + i.

To find a polynomial with these zeros, we can use the fact that if a polynomial has zeros a, b, and c, then the polynomial can be written as P(x) = (x - a)(x - b)(x - c).

So in this case, the polynomial can be written as:

P(x) = (x - (2 + i))(x - (2 - i)) [using the fact that the zeros are 2 + i and 2 - i]

Now let's simplify this expression:

P(x) = (x - 2 - i)(x - 2 + i)

Using the FOIL method to multiply these binomials, we get:

P(x) = (x - 2)(x - 2) + (x - 2)(i) + (x - 2)(-i) + (i)(-i)

Simplifying further:

P(x) = (x^2 - 4x + 4) + (ix - 2i) + (-ix + 2i) + (-i^2)

Since i^2 = -1, this simplifies to:

P(x) = x^2 - 4x + 4 + ix - 2i - ix + 2i + 1

Finally, combining like terms, we can write the polynomial as:

P(x) = x^2 - 4x + 5

So, the polynomial with integer coefficients that satisfies the given conditions is P(x) = x^2 - 4x + 5.