If a, b and c are zeros of a polynomial function, we can present this function as:
[tex]f(x)=a(x-a)(x-b)(x-c)[/tex]
Therefore we have:
[tex]f(x)=a(x-(-5))(x-(-2))(x-(-1))=a(x+5)(x+2)(x+1)[/tex]
We know f(-3) = 4. Substitute x = -3 to the equation of the function f:
[tex]a(-3+5)(-3+2)(-3+1)=4\\\\a(2)(-1)(-2)=4\\\\4a=4\qqaud|:4\\\\a=1[/tex]
Therefore we have:
[tex]f(x)=1(x+5)(x+2)(x+1)=(x+5)(x+2)(x+1)[/tex]
Use distributive property:
[tex]f(x)=(x^2+2x+5x+10)(x+1)=(x^2+7x+10)(x+1)\\\\=x^3+x^2+7x^2+7x+10x+10\\\\\boxed{f(x)=x^3+8x^2+17x+10}[/tex]
Using the Factor Theorem, the polynomial function is given by:
[tex]f(x) = x^3 + 8x^2 + 17x + 10[/tex]
The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, ..., x_n[/tex] is given by:
[tex]f(x) = a(x - x_1)(x - x_2)...(x - x_n)[/tex]
In which a is the leading coefficient.
In this problem, we have zeros -5, -2 and -1, thus [tex]x_1 = -5, x_2 = -2, x_3 = -1[/tex], and:
[tex]f(x) = a(x - x_1)(x - x_2)(x - x_3)[/tex]
[tex]f(x) = a(x + 5)(x + 2)(x + 1)[/tex]
[tex]f(x) = a(x^2 + 7x + 10)(x + 1)[/tex]
[tex]f(x) = a(x^3 + 8x^2 + 17x + 10)[/tex]
f(-3) = 4 means that when [tex]x = -3, y = 4[/tex], and this is used to find a.
[tex]4 = a((-3)^3 + 8(-3)^2 + 17(-3) + 10)[/tex]
[tex]4a = 4[/tex]
[tex]a = \frac{4}{4}[/tex]
[tex]a = 1[/tex]
Thus, the function is:
[tex]f(x) = x^3 + 8x^2 + 17x + 10[/tex]
A similar problem is given at https://brainly.com/question/24380382
