Respuesta :
ANSWER
[tex]f(x)=-\frac{7}{12}x^3+\frac{49}{12}x+\frac{7}{2}[/tex]EXPLANATION
If x1, x2, x3, ..., xn are the zeros of a polynomial P, then the polynomial can be written as the product of the factors,
[tex]P(x)=(x-x_1)(x-x_2)(x-x_3)\ldots(x-x_n)[/tex]In this problem we have a third-degree polynomial function, so it has 3 zeros and thus 3 factors,
[tex]f(x)=a(x+2)(x+1)(x-3)[/tex]We have to find a knowing that the function has to pass through point (2, 7). This means that when x = 2, f = 7,
[tex]f(2)=7[/tex]Replace into the function,
[tex]7=a(2+2)(2+1)(2-3)[/tex]Solve the parenthesis,
[tex]7=a(4)(3)(-1)[/tex]Multiply,
[tex]7=a(-12)[/tex]And solve for a by dividing both sides by -12,
[tex]\begin{gathered} \frac{7}{-12}=\frac{a(-12)}{-12} \\ a=-\frac{7}{12} \end{gathered}[/tex]Hence, the function is
[tex]f(x)=-\frac{7}{12}(x+2)(x+1)(x-3)[/tex]Next, we have to multiply the factors to obtain the function in standard form. Multiply the first two,
[tex]f(x)=-\frac{7}{12}(x\cdot x+2x+1x+2\cdot1)(x-3)[/tex][tex]f(x)=-\frac{7}{12}(x^2+3x+2)(x-3)[/tex]Then multiply by the last factor,
[tex]f(x)=-\frac{7}{12}(x^2\cdot x+3x\cdot x+2\cdot x-3\cdot x^2-3\cdot3x-3\cdot2)[/tex][tex]f(x)=-\frac{7}{12}(x^3+3x^2+2x-3x^2-9x-6)[/tex]Add like terms,
[tex]f(x)=-\frac{7}{12}\lbrack x^3+(3x^2-3x^2)+(2x-9x)-6\rbrack[/tex][tex]f(x)=-\frac{7}{12}(x^3-7x-6)[/tex]And finally, distribute the coefficient,
[tex]f(x)=-\frac{7}{12}x^3+\frac{7}{12}7x+\frac{7}{12}6[/tex][tex]f(x)=-\frac{7}{12}x^3+\frac{49}{12}x+\frac{7}{2}[/tex]This is the polynomial function with zeros -2, -1 and 3 that passes through point (2, 7)
