Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Fourth-degree and a single zero of 3

Question

Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms.
Fourth-degree and a single zero of 3

Answer 1

Hi Elizabeth.

 

If you have a fourth degree polynomial whose only 0 is 3, then you can say your polynomial takes the form of:

 

f(x) = (x-3)(x-3)(x-3)(x-3)

 

= (x² - 6x + 9)(x² - 6x + 9)

 

= x⁴ - 6x³ + 9x² -6x³ + 36x² - 54x + 9x² - 54x + 81

 

= x⁴ - 12x³ + 54x² - 108x + 81

 

There's your fourth degree polynomial. It has a "quadruple root" of 3. No other zeros (either real or imaginary) exist in this function.

 

Hope this helps.

Answer 2

Characterizing all solutions to this problem. This one is a more advanced solution than what the question is looking for.

 

If 3 is a root then (x-3) divides f. The resulting degree 3 polynomial has real coefficients because f has real coefficients. It has a real root because its degree is odd and roots with imaginary portion come in pairs if the coefficients are real (corollary to the fundamental theorem of algebra). Therefore it has a real root of 3, so (x-3)² divides f. The resulting degree 2 polynomial has real coefficients because f has real coefficients. It could have 3 as a root with multiplicity 2, which is the solution already given, (x-3)⁴, or it could have any conjugate pair of roots with non-zero imaginary portion, say i and -i, (x+i)(x-i) = x²+1.

 

The solution varies by at most a constant multiple from the ones given, so is of the form a(x-3)²(x²+bx+c) for some real a,b,c with a ≠0 and b²-4c < 0, or it is of the form a(x-3)⁴, for some real a.

 

You can see this other variety of solution by going to desmos.com and graphing (x-3)²(x²+4), for example.