Question

Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the remaining zeros must be a real number. Write down one of the missing zeros.

Answer 1

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form $ax^2+bx+c$. The cuadratic formula tells us that the solutions are of the form

$x = \frac{-b\pm \sqrt[]{b^2-4ac}}{2a}$.

Note that square root, over the reals, tells us that the are only real solutions if $b^2-4ac \geq 0$. If that is not the case, say it's negative, the solution are complex. Then, the solutions are of the form

$x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}$. NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.