Question

The function f(x) = -x³-7x2²-7x+15 has zeros located at-5, -3, 1. Verify the zeros of f(x) and explain how you verified them. Describe the end behavior of the function.

Answer 1

Explanation:

`f(x)=-x^3-7x^2-7x+15\\Prove:\ x_1=-5\ \ \ \ x_2=-3\ \ \ \ x_3=1\ at\ f(x)=0\\\\-x^3-7x^2-7x+15=0`

Multiply both parts of the equation by -1:

`x^3+7x^2+7x-15=0\\x^3+5x^2+2x^2+7x-15=0\\x^2*(x+5)+(2x^2+7x-15)=0\\x^2*(x+5)+(2x^2+10x-3x-15)=0\\x^2*(x+5)+(2x*(x+5)-3*(x+5))=0\\x^2*(x+5)+(x+5)*(2x-3)=0\\(x+5)*(x^2+2x-3)=0\\x+5=0\\x=-5\\x^2+2x-3=0\\x^2+3x-x-3=0\\x*(x+3)-(x+3)=0\\(x+3)*(x-1)=0\\x+3=0\\x=-3\\x-1=0\\x=1`