Construct a polynomial function with the following properties: fifth degree, 3 is a zero of multiplicity 2, -4 is the only other zero, leading coefficient is 2.

Answer 1

Since you want a fifth degree polynomial with with the leading coefficient of 2 we know it should look something like...

f(x) = 2x⁵ + ax⁴ + bx³ + dx² + cx + e

We also know something about roots and their multiplicity...

(x - 3)²(x - 4) represents the component of this polynomial based on the roots known...

Let's now marry the information together after multiplying out the result above...

(x - 3)²(x - 4) = (x² -6x + 9)(x - 4) = x³ - 4x² -6x² + 24x + 9x - 36 = x³ - 10x² + 33x - 36

So f(x) must be a quadratic times the above cubic to make a fifth degree polynomial.

f(x) = (ax² + bx + c)(x³ - 10x² + 33x - 36)

f(x) = ax⁵ - 10ax⁴ + 33ax³ - 36ax² + bx⁴ - 10bx³ + 33bx² - 36bx + cx³ - 10cx² + 33cx - 36c

f(x) = ax⁵ + (b-10)x⁴ + (33a - 10b + c)x³ + (33b - 36a -10c)x² + (33c - 36b)x - 36c

Solving for a --> a = 2 [ requirement from problem ]

Solving for c --> -36c = -36 --> c = 1 [ requirement from roots ]

The coefficient on the x³ gives us...

33a - 10b + c = 1 --> 33(2) - 10b + 1 = 1 --> b = -66/10 = -33/5

So the polynomial amounts to plugging in a,b,c into the bolded equation for f(x)