Question

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

Answer 1

Answer:

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

Step-by-step 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:

$11=a(2+3)(2+2)(2-1)\\ \\11=a\cdot 5\cdot 4\cdot 1\\ \\20a=11\\ \\a=\dfrac{11}{20}=0.55\\ \\P(x)=0.55(x+3)(x+2)(x-1)$