Answer:
[tex]\large\boxed{f(x)=x^2+16x+60}[/tex]
Step-by-step explanation:
The factored form of a quadratic function:
[tex]f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)[/tex]
x₁, x₂ - the zeros
We have x₁ = -10 and x₂ = -6.
Substitute:
[tex]f(x)=(x-(-10))(x-(-6))=(x+10)(x+6)[/tex]
Use FOIL: (a + b)(c + d) = ac + ad + bc + bd
[tex]f(x)=(x)(x)+(x)(6)+(10)(x)+(10)(6)\\\\f(x)=x^2+6x+10x+60\\\\f(x)=x^2+16x+60[/tex]
