Question

Find the equation for a polynomial f(2) that satisfies the following:Degree 3- Zero at x = 2• Zero at x = -2• Zero at x = -3y-intercept of (0,4)

Answer 1

In general, a polynomial of degree 3 is given by the next expression

`f(x)=a_3x^3+a_2x^2+a_1x+a_0=c(x-b_1)(x-b_2)(x-b_3)`

Where b_1, b_2, and b_3 are the roots of the polynomial.

Therefore, in our case,

`\begin{gathered} x=2\to x-2=0,x=-2\to x+2=0,x=-3\to x+3=0 \\ f(x)=c(x-2)(x+2)(x+3) \end{gathered}`

We need to find the value of c.

For this, notice that the f(0)=4

`\begin{gathered} f(0)=4 \\ \Rightarrow c(-2)(2)(3)=4 \\ \Rightarrow c=(4)/(-2\cdot2\cdot3)=(4)/(-12)=-(1)/(3) \end{gathered}`

Therefore,

The answer is f(x)=(-1/3)(x-2)(x+2)(x+3), which is equivalent to -x^3/3-x^2+4x/3+4