Question

Third degree, with zeros of -3,-2, and 1, and passes through the point (2,11)

P(x)=

Answer 1

`P(x)=0.55(x+3)(x+2)(x-1)`

Explanation:

If a polynomial function is third degree function and has zeros at -3,-2, and 1, then its equation is

`P(x)=a(x-(-3))(x-(-2))(x-1)\\ \\P(x)=a(x+3)(x+2)(x-1)`

It passes through the point (2,11), so the coordinates of the point satisfy the equation of the function:

`11=a(2+3)(2+2)(2-1)\\ \\11=a\cdot 5\cdot 4\cdot 1\\ \\20a=11\\ \\a=(11)/(20)=0.55\\ \\P(x)=0.55(x+3)(x+2)(x-1)`