Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point. Degree 3 Zeros. −4, 1 + 3 i Solution Point 3 f(−2) = 24 f(x) =

Answer 1

f(x) = a(x + 4)(x -1 -3i)(x - 1 + 3i) = a(x + 4)((x - 1)² - (3i)²) = a(x + 4)(x² - 2x + 10)

f(- 2) = a(- 2 + 4)(4 + 4 + 10) = 36a = 24; a = 2/3

f(x) = 2/3(x + 4)(x² - 2x + 10) = 2/3x³ + 4/3x² + 4/3x + 80/3

Answer 2

Ashley,

Here is an example that I hope will help:

roots are 5 and 2 + 7i and degree is 3 with f(0) = 10

Complex roots always are "conjugate pairs" when working with polynomials--I hope you have discussed this in class-- so we have

p(x) = (x - 5)(x - (2 + 5i))(x - (2 - 5i)) + C

then you multiply part of the the polynomial to get p(x) = (x - 5)(x² - 4x + 29) + C

then make use of f(0) = 10

10 = (0 - 5)(0² - 0 + 29) + C

10 = -145 + C

C = 155

Then you can multiply the remaining part of the polynomial and simplify.

Hope this helps--Andrew

Answer 3

The zero - 4 implies a factor of P(x) that is (x+4)

The zeroes, 1 +/- 3i implies a factor of (x² + the product of the conjugates) In this case the product is 10

Now just solve for the constant c so that P(x) = c(x+4)(x²+10) fulfills the last condition