Form a polynomial, f(x), with real coefficients having the given degree and zeros.
Degree: 4; zeros: 6i and 7i
I completely don't know what to do with this problem... if someone can solve and give a good explanation, I'd appreciate it.
Thanks.
Because the coefficients of f(x) are REAL,
f(x) has zeros: 6i and -6i, 7i and -7i.
For example,
f(x)=(x-6i)(x+6i)(x-7i)(x+7i)= (x^2+36)(x^2+49)
To form the polynomial, we need to use the given zeros and the fact that the polynomial has real coefficients.
Since the zeros are complex numbers, we know that their conjugates will also be zeros of the polynomial. That means the zeros of the polynomial are:
6i, -6i, 7i, -7i
To form the polynomial, we can use the fact that the product of a binomial and its conjugate is always a real number. For example, (x - a)(x - b) will always produce a real number polynomial when a and b are complex conjugates.
For the complex zeros 6i and -6i, their corresponding binomial factors are (x - 6i) and (x + 6i). Similarly, for the complex zeros 7i and -7i, their corresponding binomial factors are (x - 7i) and (x + 7i).
Now, we can multiply these factors together to get the polynomial:
f(x) = (x - 6i)(x + 6i)(x - 7i)(x + 7i)
To simplify this expression, we can use the difference of squares formula as follows:
(x - 6i)(x + 6i) = x^2 - (6i)^2 = x^2 + 36
(x - 7i)(x + 7i) = x^2 - (7i)^2 = x^2 + 49
Now we can substitute these simplified expressions back into the polynomial:
f(x) = (x^2 + 36)(x^2 + 49)
Expanding this expression gives us the polynomial with the given zeros and real coefficients:
f(x) = x^4 + 49x^2 + 36x^2 + 1764
Combining the like terms, we get the final polynomial:
f(x) = x^4 + 85x^2 + 1764
So, the polynomial with degree 4 and zeros 6i and 7i is f(x) = x^4 + 85x^2 + 1764.