First we know f(x) is a 3rd degree polynomial => it has the following form.
[tex]f(x) = ax^3 + bx^2 + cx + d, \ a, b, c, d \in \mathbb{R}[/tex]
There is however a very famous theorem, namely ''The fundamental Theorem of Algebra'' which states that any polynomial equation of degree n, has n roots/zeroes in the real and imaginary set.
But we also know if r is a root of a polynomial than the binomial (x-r) is a factor of the original equation, these 2 statements combined give rise to the following:
[tex]let \ P(X) \ be \ a \ polynomial \ in \ \mathbb{R}[x] \ of \ degree \ n:\\\\$P(X) \ is \ a \ polynomial \ of \ degree \ n \ with \ real \ coeficients, where n \in \mathbb{N}.\\$Combined with the previous statement $=> \\P(X) = (x-r_1)(x-r_2)(x-r_3)...(x-r_n), $ where $r_i $ are the roots / zeroes.[/tex]
Using the following means that in your problem the polynomial is constructed by taking the product of the linear factors / (x-r) where r are the roots. One such factor for each root which yields:
[tex]f(x) = a(x+8)(x-1)(x-6) \ f(-4) = 7.\\$Plugging -4 into x to find a :$\\f(-4) = a(-4+8)(-4-1)(-4-6) = 200a = 7 => a = \frac{7}{200}.\\$Thus $P(X) = \frac{7}{200}(x+8)(x-1)(x+6)[/tex]
