The quartic function whose only real zeros are -3 and 2 is [tex]p(x) = x^{4}+2\cdot x^{3}-11\cdot x^{2}-12\cdot x +36[/tex].
By definition of polynomial, we can define it as a finite product of binomials described below:
[tex]p(x) = \Pi\limits_{i= 1}^{n} (x-r_{i})[/tex] (1)
Where:
Based on the information from statement, we know that [tex]n = 4[/tex], [tex]r_{1} = r_{2} = -3[/tex] and [tex]r_{3} = r_{4} = 2[/tex] and we get the following expression:
[tex]p(x) = (x+3)^{2}\cdot (x-2)^{2}[/tex]
[tex]p(x) = (x^{2}+6\cdot x + 9) \cdot (x^{2}-4\cdot x +4)[/tex]
[tex]p(x) = x^{2}\cdot (x^{2}-4\cdot x + 4) +6\cdot x \cdot (x^{2}-4\cdot x + 4) + 9\cdot (x^{2}-4\cdot x + 4)[/tex]
[tex]p(x) = x^{4}-4\cdot x^{3}+4\cdot x^{2}+6\cdot x^{3}-24\cdot x^{2}+24\cdot x +9\cdot x^{2}-36\cdot x+36[/tex]
[tex]p(x) = x^{4}+2\cdot x^{3}-11\cdot x^{2}-12\cdot x +36[/tex]
The quartic function whose only real zeros are -3 and 2 is [tex]p(x) = x^{4}+2\cdot x^{3}-11\cdot x^{2}-12\cdot x +36[/tex].
We kindly invite to check this question on polynomials: https://brainly.com/question/11536910
