PreCalc questions

Question

Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point. Degree:3, zeroes: -7, 1+sq. root of 3i, Solution points f(-2)=60

Answer 1

The key to this is to know that anytime you have a complex solution, like 1 + √3i, you automatically and always will have the conjugate, 1 - √3i.

Having zeros of -7, 1 + √3i, and 1 - √3i means the FACTORS of the polynomial will be:

(x + 7)[x - (1 + √3i)][ x - (1 - √3i)] or (x + 7)(x - 1 - √3i)(x - 1 + √3i) however, the polynomial MAY be multiplied by a constant term so we must write it as:

P(x) = a(x + 7)(x - 1 - √3i)(x - 1 + √3i)

At this point, it's better to multiply out the two complex conjugates because the product will have real number coefficients and it's easier to work with.

Use whatever method you normally use to multiply (x - 1 - √3i)(x - 1 + √3i) and you'll get:

(x - 1 - √3i)(x - 1 + √3i) = x² - 2x + 4

So P(x) = a(x + 7)(x² - 2x + 4)

Using the point (-2, 60), we can solve for "a":

60 = a(-2 + 7)((-2)² - 2(-2) + 4)

60 = a(5)(12)

60 = 60a

a = 1

So P(x) = (x + 7)(x² - 2x + 4)

If you want (or perhaps you should), multiply the factors to get the polynomial in standard form.