Answer:
[tex]\large\boxed{f(x)=x^3+3x^2-28x-60}[/tex]
Step-by-step explanation:
If a, b and c are the zeros of a cubic function, then a function equation can have a form:
[tex]f(x)=(x-a)(x-b)(x-c)[/tex]
We have the zeros: -2, 5 and -6. Then:
[tex]f(x)=(x-(-2))(x-5)(x-(-6))=(x+2)(x-5)(x+6)[/tex]
Use FOIL (a + b)(c + d) = ac + ad + bc + bd:
[tex]=\left[(x)(x)+(x)(-5)+(2)(x)+(2)(-5)\right](x+6)\\\\=(x^2-5x+2x-10)(x+6)\\\\=(x^2-3x-10)(x+6)\\\\=(x^2)(x)+(x^2)(6)+(-3x)(x)+(-3x)(6)+(-10)(x)+(-10)(6)\\\\=x^3+6x^2-3x^2-18x-10x-60\\\\=x^3+(6x^2-3x^2)+(-18x-10x)-60\\\\=x^3+3x^2-28x-60[/tex]
