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Question

Find a fourth-degree polynomial with integer coefficients that has    and    as zeroes, such that 

Answer 1

In any polynomial with integer coefficients, an imaginary root will appear with its complex conjugate. What this means is that since -2i and -3i are zeroes, so are 2i and 3i. Thus, the polynomial is of the form, k(x+2i)(x-2i)(x+3i)(x-3i), where k is any integer. But remember that for any a, (x-a)(x+a) = x^2 -a^2. Thus, (x+2i)(x-2i) = x^2 -(2i)^2 = x^2 -(-4) = x^2 +4. Similarly, you can check that (x+3i)(x-3i) = x^2 + 9. So, we have the our polynomial is of the form k(x^2+4)(x^2+9).

Now, we will plug in -2 for x, since we are given f(-2)=312. We have k((-2)^2+4)((-2)^2+9) = 312, or, k(4+4)(4+9) = 312, or k(8)(13) = 312.

Now, we have that 312=2*156= 2*2*78=2*2*2*39=2*2*2*(3*13) = 3(8)(13). By replacing 312 with 3(8)(13), we have that k(8)(13) = 3(8)(13). It is easy to see that k=3.

Thus, our polynomial is 3(x^2+4)(x^2+9).