Question

Construct a polynomial function with the stated properties. Reduce all fractions to lowest terms. Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,10).

Answer 1

Answer:

$\Large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}$

Step-by-step explanation:

Hello,

Based on the indication, we can write this polynomial as below, k being a real number that we will have to identify (degree = 3 and we have three zeroes -3, -1, and 2).

   $\Large \boxed{k(x+3)(x+1)(x-2)}$

We know that the point (1,10) is on the graph of this function, so we can say.

$k(1+3)(1+1)(1-2)=10}\\\\4*2*(-1)*k=10\\\\-8k=10\\\\k=\dfrac{10}{-8}=-\dfrac{5}{4}$

Then the solution is:

$\large \boxed{-\dfrac{5}{4}(x+3)(x+1)(x-2)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you