Polynomials precal

Question

Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.

degree=4

Zeros=-3,1,i

solution point f(0)=-12

Answer 1

You should have studied the Fundamental Theorem of Algebra which says (in essence) the degree of the polynomial equals the number of zeros. So 4th degree means there are 4 roots. But there are only 3 given. What is the other root? You should also have studied that imaginary roots ONLY come in pairs that are conjugates. So, since "i" is a zero, then "-i" must also be a zero. So the zeros are -3, 1, i, and -1.

Additionally, you should have studied that if a number is a zero, it can be written as a factor (x - #) where # is the zero.So since the zeros are -3, 1, i, and -1, then the factors are (x + 3)(x - 1)(x - i)(x + i). We can multiply these together to get a polynomial that you're probably more used to seeing. Using the FOIL method, (x + 3)(x - 1) = x² + 2x -3 and (x - i)(x + i) = x² +1. Then we multiply (x² + 2x -3)(x² + 1) to get x⁴ + 2x³ - 2x² + 2x - 3. So the resulting function would be f(x) = x⁴ + 2x³ - 2x² + 2x - 3

Now we need to make sure the solution point works. Since f(0)=-12, that means the point (0, -12) is a solution. But when we plug in "0", to get f(0), we get -3 [f(0) = 0⁴ + 2(0)³ - 2(0)² + 2(0) - 3 = -3]. How do we fix this? We need to remember that when we wrote the function in factored form [f(x) = (x + 3)(x - 1)(x - i)(x + i)], we could have also written it as f(x) = a(x + 3)(x - 1)(x - i)(x + i) where "a" is the vertical stretch factor. So let's re-write our function as f(x) = a(x⁴ + 2x³ - 2x² + 2x - 3) and solve for "a" using the point (0, -12). So, -12 = a(0⁴ + 2(0)³ - 2(0)² + 2(0) - 3) or -12 = -3a meaning a = 4. So the final function would be: f(x) = 4(x⁴ + 2x³ - 2x² + 2x - 3) or, if you want to multiply it out, f(x) = 4x⁴ + 8x³ - 8x² + 8x - 12