Respuesta :
Answer:
f(x) = x³ + (9/2)x² - 10x - 6
Explanation:
A polynomial with zeros at b, c, d has the following form
f(x) = a(x - b)(x - c)(x - d)
Where a is the leading coefficient.
In this case, a = 1, and the zeros are -1/2, 2, and -6. Then, we can write the polynomial as follows
[tex]\begin{gathered} f(x)=1(x-(-1/2))(x-2)(x-(-6_{})) \\ f(x)=(x+\frac{1}{2})(x-2)(x+6_{}) \end{gathered}[/tex]To write it in standard form, we need to expand the expression, so
[tex]\begin{gathered} f(x)=(x(x)+x(-2)+\frac{1}{2}(x)+\frac{1}{2}(-2))(x+6) \\ f(x)=(x^2-2x+\frac{1}{2}x-1)(x+6) \\ f(x)=(x^2-\frac{3}{2}x-1)(x+6) \\ f(x)=x^2(x)+x^2(6)-\frac{3}{2}x(x)-\frac{3}{2}x(6)-1(x)-1(6) \\ f(x)=x^3+6x^2-\frac{3}{2}x^2-9x-x-6 \\ f(x)=x^3+\frac{9}{2}x^2-10x-6 \end{gathered}[/tex]Therefore, the polynomial in standard form is
f(x) = x³ + (9/2)x² - 10x - 6
